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History

Business objects is a French enterprise software company, specializing in business intelligence (BI). The company claimed more than 46,000 customers worldwide in its final earnings release. Its flagship product is BusinessObjects XI, with components that provide performance management, planning, reporting, query and analysis and enterprise information management.
1990    Business Objects is founded in Paris, France.
1991    France Telecom becomes the first customer after viewing a prototype of Business Objects’ initial product offering.
1994    Business Objects launched BusinessObjects v3.0 and goes public on NASDAQ – the first European company in history to do so.
1995    Business Objects is the first software company to focus on enterprise-scale business intelligence (BI) deployments.
1996    Business Objects enters OLAP market and launches BusinessObjects v4.0.
Bernard Liautaud is named one of Business Week’s “Hottest Entrepreneurs of the Year.”
1997    BusinessObjects Web Intelligence is introduced.
1999    Business Objects is publicly listed in France on the Premier Marche.
2000    Business Objects delivers the industry’s first interactive mobile business intelligence solution.
2001    Business Objects opens offices in San Jose, California.
signs an OEM and reseller agreement to bundle Crystal Reports.
2002    Business Objects acquires Blue Edge Software and Acta Technologies.
Bernard Liautaud named to Business Week’s “Stars of Europe.”
Business 2.0 names Business Objects as one of the “100 Fastest Growing Tech Companies.”
Leading direct retailer Lands’ End deployed one of the first executive dashboards, resulting in a 33% reduction in lost sales over the year.
2003    Business Objects acquires Crystal Decisions and releases Dashboard Manager, BusinessObjects Enterprise 6, as well as BusinessObjects Performance Manager.
2004    Business Objects launches Crystal v10 and BusinessObjects v6.5.
2005    Revenues reaches more than a billion US dollars.
John Schwarz joins Business Objects as chief executive officer.
BusinessObjects XI and Crystal Xcelsius are launched.
Business Objects acquires SRC Software, Medience, and Infommersion.
2006    Business Objects acquires Firstlogic, Inc., Armstrong Laing Limited (ALG), and nSite.
Business Objects releases crystalreports.com and BusinessObjects Data Quality XI.
2007    Business Objects acquires Cartesis.
2008    acquires Business Objects to lead emerging market for business performance optimization.
2002 More birthday cake: Thirty years after its foundation,  is the third-largest independent software provider in the world and a paragon of the German economy. The  brand stands for high-quality business software.

A rising star: ‘s portfolio is not the only thing witnessing constant growth. The company’s workforce also increases to around 29,000 by the end of 2002. Approximately 1,300 employees move into the new star-shaped building that has opened directly adjacent to ‘s headquarters in Walldorf.

New blood at the top: The Executive Board bolsters its ranks with Shai Agassi, who assumes responsibility for new technologies, and Léo Apotheker, who takes over global sales.
2003

End of an era: Hasso Plattner resigns from the Executive Board and is elected chairman of the Supervisory Board. Plattner is the final  co-founder to leave the company’s management team, but remains with  in an advisory role. Upon taking his leave, the nautical enthusiast receives a special gift – a sail signed by all of ‘s employees, which they hope will carry him to further success.

Technology of the future: What began in the “new economy” as my.com and evolved to my technology reaches a new pinnacle in  NetWeaver. This technology enables  to offer fast, open, and flexible business applications that support end-to-end business processes – no matter whether they are based on systems from  or other providers.

Global developments: Labs China marks the ninth opening of a development location outside of Walldorf. This and the other research centers in India, Japan, Israel, France, Bulgaria, Canada, and the United States help  convert IT expertise into business utility for its customers. The company now employs around 30,000 employees, approximately 17,000 of whom work outside of Germany.
2004

Major success: brings the first version of  NetWeaver to market. The response to this new integration and application platform is overwhelming. By the end of the year, well over 1,000 customers acquire the product, with even more on the way. Meanwhile, more than 24,000 total customers are running 84,000  software installations in over 120 countries.

Joining forces: announces its intention to acquire the remaining shares of its consulting subsidiary  SI and to merge the company into its corporate group. This move strengthens ‘s global portfolio of strategic IT consulting and integration services and makes  the go-to provider of many customers – especially those in Germany, Switzerland, and the United States.

A clear vision: plans its future around the concept of enterprise service-oriented architecture (enterprise SOA). According to CEO Henning Kagermann,  will make all of its business applications service-based in the medium term to provide its customers with the most flexibility possible. In doing so,  sets the standard for the rest of the market.

Strong statistics: Under Kagermann’s leadership, quarterly revenue gains and a constantly increasing market share keep  at the head of the pack in the rejuvenated IT market. Business Week names Kagermann one of the 25 best business managers in Europe, praising his customer-oriented corporate philosophy.
2005

Excellence recognized: A study conducted on behalf of the business magazine Capital names  Germany “Best employer of 2005″ among other companies with 5,000 employees or more. “I accept this award with pride and gratitude on behalf of the more than 32,000 people who work at . For us as a company, it will mainly serve as motivation for the future. After all, ‘s success will continue to depend on the skills, drive, and dedication of our current and future employees,” declares  Executive Board member Claus Heinrich upon receiving the award from Wolfgang Clement, Germany’s Federal Minister of Economics and Technology.

Something cooking in the east: In February,  officially opens its new research and development facility in the Hungarian capital of Budapest.  Labs Hungary and its some 50 developers join ‘s global network of research locations, which incorporates the brightest 
minds in IT all over the world.

Organic growth: The year 2005 is marked by a series of acquisitions. While its competitors initiate their own major takeovers,  focuses on organic growth by acquiring smaller companies whose specific solutions augment its portfolio in sensible ways. These companies include the two retail providers Triversity and Khimetrics.

Impressive numbers: The company’s software license revenues increase by 18%, and it records particularly high rates of growth in the Americas. The more than 35,800  employees around the world generate total revenues of €8.5 billion.
2006

International recognition: once again garners numerous employer accolades. Along with  Germany,  Austria,  Chile,  Andina y del Caribe,  Mexico, and  Region Sur win the “Great Place to Work” award from the institute of the same name. In addition,  Labs India receives the distinction “Recruiting and Staffing Best in Class” from the Indian Institute of Management Studies & Research for its innovative methods in workforce planning and management.

Harmonious partnership: and Microsoft introduce Duet, the first product of the two companies’ joint efforts in development, support, sales, and marketing. This software enables users to quickly and easily integrate Microsoft Office and -supported business processes. The partners sell 200,000 licenses in just the first three months.

Onward to further success: At the first PHIRE event of the year in Orlando, Florida,  announces the general release of its flagship application,  ERP. ” ERP constitutes the foundation of ‘s industry solutions and a springboard to enterprise SOA,” says  Executive Board member Léo Apotheker.

Good news from the midmarket: noticeably expands its share of the midmarket with the solutions  All-in-One and  Business One. In June, the company announces that the latter solution has gained its 10,000th customer.  also has generates around 30% of its €3.1 billion in total software licensing revenues from companies with fewer than 2,500 employees.
2007

Seizing new opportunity: When an acquisition makes sense in enhancing its product portfolio,  does not shy away from the investment required. The company buys Pilot Software – a private California provider of strategy management software – as well as Yusa, OutlookSoft, Wicom, and MaXware.  also announces its intention to purchase Business Objects, a company specializing in business intelligence applications.

Restructure of the Executive Board: Leo Apotheker is named ‘s deputy CEO at the end of March.  also forms an Executive Council, which comprises corporate officers who share responsibilities for market and product strategies and report to the Executive Board. This restructuring follows Shai Agassi’s departure from the Executive Board.

Another year, another employer accolade: In its 35th year,  takes home the title of “Germany’s Best Employer” for the third time in the annual “Great Place to Work” awards. The company also receives a special prize for its comprehensive health management program.  Labs India also ranks eighth among other employers in India.

Head start in the midmarket: As part of a special event in New York City,  reveals  Business By Design, a product designed specifically for small businesses and midsize companies.  CEO Henning Kagermann states that this new offering “represents ‘s ambitious attempt to create an all-new solution for an untapped market.”
2008

In the bag: successfully completes its acquisition of Business Objects. Purchasing the French provider of business intelligence solutions expands ‘s software portfolio and makes it the market leader in business software, enterprise performance management, and business intelligence.

The choice of skilled employees: For the fourth time,  is named “Germany’s Best Employer” among companies with at least 5,000 employees. The company also receives numerous awards in other countries, including China, Bulgaria, Denmark, India, Japan, and Mexico.

A global focus: The  Supervisory Board names Léo Apotheker co-CEO alongside Henning Kagermann. Two longstanding members, Peter Zencke and Claus Heinrich, resign from the Executive Board. The board then welcomes Ernie Gunst, Bill McDermott, and Jim Hagemann Snabe, whose international backgrounds will enrich ‘s executive management.

Sustainable business: proves its commitment to sustainable business practices, releasing its first Sustainability Report. As the leader in its market,  is in a unique position to provide information technology that helps companies and organizations of all sizes improve their track records and achieve long-term sustainability.
2009

Difficult times: With the effects of the global financial crisis having reached the real economy in 2008, the business world faces its own plight. Susceptible to the situation at hand,  initiates personnel cutbacks and other cost-saving measures. As of Q3 2009,  still employs some 47,800 people. Meanwhile, the company supports its customers with special programs designed to help them emerge from the crisis with the strength to succeed. Thanks to these programs and its cutbacks,  is able to improve its operating margin despite the difficult circumstances.

Another business milestone: At a launch event at its offices in New York City,  unveils its  Business Suite 7 software, which is designed to help businesses optimize their performance and reduce IT costs. A condensed ramp-up phase enables the first customers to go live with the software in March. In early May, the next generation of the suite is released to the rest of the world.

Passing the torch:  After 27 years at the company – including 18 years on the Executive Board – Henning Kagermann bids farewell to .  Léo Apotheker becomes the company’s sole CEO. In his inaugural address to ‘s employees in June, he stakes out a clear new path for the company, including his plans for ‘s future, its purpose, and the associated values.

In it for the long haul: demonstrates how important it considers social involvement by supporting PlaNet Finance, an international non-profit organization that aids microfinance institutions (MFIs).  and PlaNet Finance aim to optimize the microfinance sector with a combination of financing, new technologies, and expanded value chains. In advance of the 15th United Nations Climate Change Conference in Copenhagen (COP15),  also offers its assistance to the U.N.’s Hopenhagen initiative. Last but not least,  employees continue to volunteer their time in support of social projects all over the world.
2010

Dynamic duo: In February, the Supervisory Board names Bill McDermott and Jim Hagemann Snabe co-CEOs of the company. Chief technology officer Vishal Sikka also joins the Executive Board. Angelika Dammann follows Sikka in July, becoming the first woman to serve on the Executive Board as she assumes responsibility for global human resources and labor relations.

Business Objects pioneered the business intelligence (BI) industry 
with the invention of the semantic layer, which shields BI users from the complexity of databases.

Business Objects is once again ushering in a new era of BI by delivering a breakthrough innovation that allows anyone to unthinkingly ask business questions, without the need to build a query or know the underlying systems or data. Intelligent Question is the industry’s first structured questioning environment and is designed to work the way people think. With Intelligent Question.  
Business Objects is the first vendor to solve this problem by delivering breakthrough ease of use. Just like the introduction of the easy-to-use Apple iPod has led to an explosion in demand for MP3 players, Intelligent Question has the potential to introduce millions of people to the benefits of business intelligence. “The majority of people consuming reports are not technical enough, laymen to create an ad hoc query,” said Philip Russom, senior manager of research and services at The Data Warehousing Institute (TDWI). “This is a problem when end-users need to discover information quickly and standard reports do not contain the answer. These people need an information discovery mechanism that’s as fast, easy, and spontaneous as a search engine, while applying structure to the answer the way parameterized reports do. BusinessObjects Intelligent Question addresses this problem by enabling people to ask an English-language question. Under the hood, Intelligent Question converts the question to an ad hoc or parameterized query, and then returns information in an easy-to-digest format. With this kind of technology, BI users can find information in their reporting systems based on their immediate needs, in a quick, self-service fashion.” When critical business questions arise, people want fast, precise answers. Unfortunately, people often do not know how to best structure their questions to ensure they receive the right information. Intelligent Question’s structured questioning environment allows people to select from a series of options to create anything from simple to very complex business questions. For example, an executive at a retail organization could ask: What are my top 15 customers, based on average weekly spend for my stores over the past 12 months? Or, a supplier for a computer hardware company could easily craft a question via an extranet to ask: Which of my parts have experienced a 10 percent increase in demand this month over last month? With Intelligent Question, as people select from the preset options, the remaining choices are automatically filtered to guarantee that the user can only create an “intelligent question.” A correct, relevant, and trusted answer is provided every time, with just one click.  BusinessObjects Intelligent Question is the first solution that will finally break down the barrier to broader BI adoption, and truly meet the needs of those users who have not experienced the benefits of BI because they were intimidated by the complexity.

Interview Questions

Business Objects Interview Questions

1. What is ‘business objects’?

BUSINESSOBJECTS is an integrated query, reporting and analysis solution for business professionals that allow them to access the data in their corporate databases directly from their desktop and present and analyze this information in a BUSINESSOBJECTS document.

It is an OLAP tool that high-level management can use as a part of a Decision Support Systems (DSS).

BUSINESSOBJECTS makes it easy to access the data, because you work with it in business terms that are familiar to you, not technical database terms like SQL.

2. What are the various Business Objects products?

User Module, Designer, Supervisor, Auditor, Set Analyzer, Info View (Web Intelligence), Business Objects Software Development Kit (SDK), Broadcast Agent etc.

3. What are the advantages of Business Objects over other DSS?

User Friendly.

Familiar Business Terms.

Graphical Interface

Drag and Drop.

Powerful reports in less time.

Enterprise wide Deployment of documents using WebI

Customized dashboards using Application foundation and Business Objects SDK.

4. How many modes are there in BO & Designer

There are 2 types: Enterprise Mode, Workgroup

5. What are Enterprise and Workgroup modes?

Designer lets you save universes in either enterprise or workgroup mode. Enterprise mode means that you are working in an environment with a repository. Workgroup mode means that you are working without a repository. The mode in which you save your universe determines whether other designers are able to access them. By default, a universe is saved in the mode in which you are already working. For example, if you launched a session in enterprise mode, any universe you save is automatically in that mode.

6. How do you save a Business Objects document which can be accessed by all users in workgroup mode?

If we want to make a universe accessible to another designer working without a repository, then click the “Save For All Users” check box in the Save as universe dialog box.

7. What is online and offline mode?

If you want a universe to be accessible in offline mode, you must firstly ensure that the

universe has been opened at least once in online mode, and that it has been saved with the Save for All Users check box selected in the Save Universe As box. To make Offline mode available:

Mode Description

Online Default mode of operation for Designer when you are working in an

environment with a repository.

Offline Mode of operation for Designer when you are not connected to a repository.

• Only available if you have previously connected in online mode.

• In offline mode you can open universes that are stored on your local computer only if those universes have been opened previously in online mode.

• You can access databases where the connection and security information are stored on your local machine (personnel and shared connections.)

• You can use offline mode when you do not have access to the repository, for example when working with a laptop off site, or when the network is not available.

8. What is a universe?

Universe provides a semantic layer between you and the database. It consists of classes and objects named in business terms. It is basically a mapping of table and the columns in the database to classes and objects respectively in the query panel.

Alternatively, It is logical mapping of data in business terms.

In the BusinessObjects User module, universes enable end users to build queries from which they can generate and perform analysis. Universes isolate end users from the complexities of the database structure as well as the intricacies of SQL syntax.

9. Can a Universe connect to multiple databases?

NO

10. How do you define universe parameters?

The first step in creating a universe is to specify its parameters. These parameters include the definition of a universe, which is comprised of: the universe name, a description of the universe, a connection to an RDBMS. You enter universe parameters from the Universe Parameters dialog box. This dialog box also lets you set up database options, external strategies, graphic options, and print settings.

11. What is a database connection?

A connection is a set of parameters that provides access to an RDBMS. These parameters include system information such as the data account, user identification, and the path to the database. Designer provides three types of connections: secured, shared, and personal.

12. What are the types of connections we use when connecting to the database?

There are three types of connections namely: – Secured, Shared and Personal.

A secured connection is used to centralize and control access to sensitive or critical data. It is the safest type of connection for protecting access to data.

A shared connection is used to access common resources such as universes or documents. Several users can thus use it.

A personal connection is specific to one user and can be used only from the computer on which it was created.

13. What are different types of joins available in Universe design? Explain each

Equi or Inner or Natural or Simple join: is based on the equality between the values in the column of one table and the values in the column of another. Because the same column is present in both tables, the join synchronizes the two tables.

Self-Join: join a table to itself i.e create a self-join to find rows  in a table that have values in common with other rows in the same table.

Theta or Non-Equi join: links tables based on a relationship other than equality between two columns. A join is a relational operation that causes two or more tables with a common domain to be combined into a single table. The purpose of joins is to restrict the result set of a query run against multiple tables.

Outer join: links two tables, one of which has rows that do not match those in the common column of the other table.

Left Outer Join: All records from first table with matching rows from second.

Right Outer Join: All records from second-named table with matching rows from left.

Full outer join: All rows in all joined tables are included, whether they are matched or not.

Shortcut join: can be used in schemas containing redundant join paths leading to the same result, regardless of direction. Improves SQL performance.

14. How do you design a universe?

The design method consists of two major phases.

During the first phase, you create the underlying database structure of your universe. This structure includes the tables and columns of a database and the joins by which they are linked. You may need to resolve loops which occur in the joins using aliases or contexts. You can conclude this phase by testing the integrity of the overall structure.

During the second phase, you can proceed to enhance the components of your universe. You can also prepare certain objects for multidimensional analysis. As with the first phase, you should test the integrity of your universe structure. Finally, you can distribute your universes to users by exporting them to the repository or via your file system.

15. What are the components of the Designer interface?

In Designer, you create a universe using three areas: the Universe pane, the Structure pane, the Table Browser.  The Universe pane displays the components of the universe from the point of view of BusinessObjects; that is the classes, objects, and conditions. The Structure pane reflects the underlying database structure of the universe including the tables, columns, and joins. The Table Browser is the component that lets you create the classes and objects of the universe from the tables and columns of a database.

16. What are classes/objects?

An object maps to data or a derivation of data in the database. For the purposes of multidimensional analysis, an object can be qualified as one of three types: a dimension, detail, or measure.

A class is a collection of objects based on business categories.

A universe is a set of classes and objects intended for a specific application or group of users.

17. What are classes?

A class is a logical grouping of objects within a universe. In general, the name of a class reflects a business concept that conveys the category or type of objects. For example, in a universe pertaining to human resources, one class might be Employees. A class can be further divided into subclasses. In the human resources universe, a subclass of the Employees class could be Personal Information. As designer, you are free to define hierarchies of classes and subclasses in a model that best reflects the business concepts of your organization.

18. What are objects?

An object is the most refined component in a universe. It maps to data or a derivation of data in the database. Using objects, end users can build queries to generate reports. The name of an object suggests a concept drawn from the terminology of a business or discipline. For a human resources manager, objects might be Employee Name, Address, Salary, or Bonus, while for a financial analyst, objects might be Profit Margin, Return on Investment, etc. For the purposes of multidimensional analysis, objects are qualified as one of three types: dimension, detail, or measure.

19. What are Dimension-Measure-Detail Objects?

When creating universes, universe designers define and qualify objects. The qualification of an object reveals how it can be used in analysis in reports. An object can be qualified as a dimension, a detail, or a measure.

A dimension object is the object being tracked; in other words, it can be considered the focus of the analysis. A dimension can be an object such as Service, Price, or Customer.

Dimension objects retrieve the data that will provide the basis for analysis in a report. Dimension objects typically retrieve character-type data (customer names, resort names, etc.), or dates (years, quarters, reservation dates, etc.)

A detail object provides descriptive data about a dimension object (or attribute of a dimension). It is always associated with a specific dimension object. However, a detail object cannot be used in drill down analysis. E.g. Address & phone number can be attributes about the customer dimension.

A measure object is derived from one of the following aggregate functions:

Count, Sum, Minimum, Maximum or average or is a numeric data item on which you can apply, at least locally, one of those functions. This type of object provides statistical information. Examples of measure objects include the following:

Revenue, unit price etc

20. What us hierarchy

Groups of related dimension objects are referred to as dimension hierarchies. An example of a dimension hierarchy is Geography, which can consist of City, Region, and Country.(Hierarchy is an ordered series of related dimensions, which can be used, in multidimensional analysis) Good examples of hierarchy are geography and time

21. How to create hierarchies in BO?

A hierarchy, which the designer sets up when creating the universe, consists of dimension objects ranked from “less detailed” to “more detailed”. The objects that belong to hierarchies are the ones you can use to define scope of analysis.

22. Can a Universe have more than 1 fact Table?

Yes. Typically a universe can have more than 1 fact table and numerous aggregated tables.

23. What are Contexts?

A context is a rule that determines which of two paths can be chosen when more than one path is possible in the database from one table to another. It helps in resolving the loops created by various joins in the universe tables.

With certain database structures, you may need to use contexts rather than aliases to resolve loops. A situation where this commonly occurs is a transactional database with multiple fact tables (“multiple stars”) that share lookup tables.

24. What are aggregated tables and how would you use it in BO Universe

Aggregate table

Aggregate tables are table which contains summarized data at different level depending on the need of reports. Imagine a fact table which contains granular data up to minute’s transaction. Now if you are developing a reports which has hour, day, week, month, quarter, year level summaries. Queries to get these summary values will scan millions of records which would in turn result in poor performance of reports. One can address this issue by creating aggregate summary table.

Possible problems of using Aggregate table

Aggregate table are good to achieve performance for high level summery queries. However if there are multiple aggregate tables which contains summary values then using aggregate table might be an issue.

Consider following example

e.g. you have two aggregate table

Table 1

AggregateID    Year    Quarter     Month    Sales Revenue

Table 2

AggregateID    Year    Quarter    Month    ServiceType    Sales Revenue

Now if you have a requirement wherein one of your report is displays year wise sales revenue whereas other report is displays Year wise ServiceType “sales revenue”. How would you tell report to use table1 for “year wise” “sales revenue” and use table 2 for another report.

Using Aggregate Table in Business Objects

Business Objects provides a wonderful function to use aggregated tables.

Aggregate_aware(). This function determines which aggregate table to use based on the attributes used in the query.

Syntax:

Aggregate_aware(<expression1>,

<expression2>,

—-

—-)

Expression = field or valid SQL expression or calculation

e.g. so formula for sales revenue might be.

Aggregate_aware (table1.salesrevenue,

table2.salesrevenue,

sum(sometable.column)

)

Notice the arrangement of column used in function. It is highest summarized to least summarized.

Now if while creating report if you use only year attribute and sales revenue. Aggregate_aware will use table1 to get the sales revenue, if used service type in report It will use table2, in all other cases it will use sum(sometable.column) which could be a fact table.

Rules to use aggregate_aware

1.If the object is at the same level of aggregation or higher, it is OMPATIBLE with the summary table.

e.g. In above example none of the table contains aggregated value for “Article” so article object will be incompatible.

You can not use aggregate_aware with “Article object”

2.If the object is at a lower level of aggregation, it is INCOMPATIBLE.

3.If the object has nothing to do with the summary table, it is INCOMPATIBLE.

25. What is Incompatibility?

The set of incompatible objects you specify determines which aggregate tables are disregarded during the generation of SQL. With respect to an aggregate table, an object is either compatible or incompatible. The rules for compatibility are as follows:

When an object is at the same or higher level of aggregation as the table, it is compatible with the table.

When an object is at a lower level of aggregation than the table (or if it is not at all related to the table), it is incompatible with the table.

26. What is the typical strategy employed in developing/maintaining/distributing Universes?

Phase 1: Break down the informational system into functional areas.

Phase 2: Analyze the information needs of users.

Phase 3: Design a conceptual schema

Design the specification of the user.

Phase 4: Create a Universe with designer.

Test the Universe with the Business Objects module.

Distribute the Universe.

Repeat the other steps for other Universes.

Phase 5:  Update and maintain the Universe.

Notify end users of changes.

27. I have Customer dimension table and a fact table with cust_to_ship_key and cust_to_bill_key. How do I get the corresponding customer names?

Create an Alias table for Customer dimension table. Join the cust_to_ship_key with a customer key of Actual customer table and join the cust_to_bill_key with customer key of the Alias.

28. What are strategies?

A strategy is a script that automatically extracts structural information from a database or flat file.

29. What are the different types of Strategies?

In Designer we can specify two types of strategies:

1)    Built-in strategies and

2)    External strategies

Built-in Strategies: Designer provides a number of default strategies which we can use. These are strategies for extracting joins, detecting cardinalities, and creating default classes and objects. Options for indicating default strategies are located in the Database tab of the Options dialog box.

External Strategies: We can also create our own strategies. Such strategies are referred to as external strategies. With an external strategy, we can specify the exact way that objects and joins are to be extracted from the database structure. The strategy we use, for example, can be a script generated from a CASE Access tool. An external strategy is specific to one RDBMS.

30. How do you specify external strategies?

With an external strategy, you can specify the exact way that objects and joins are to be extracted from the database structure. All external strategies are contained within the same text file. The name of this text file is indicated in the .prm file specific to your RDBMS. In the .prm file, the strategy file is declared as follows:  STG=[StrategyFileName] where StrategyFileName is the name of the strategy file.

An external strategy, whether for objects or for joins, is made up of the following sections:

a name and description (These are visible in the Strategies tab of the Universe Parameters dialog box.)

a type parameter: object or join

an SQL parameter or file parameter

an optional parameter that points to a connection other than the universe connection.

An external strategy can be based on SQL or a file.

31. What are the visualization options available?

Designer contains a variety of features for organizing and viewing the tables and columns in the Structure pane. Among these features are: List Mode, which adds three panes to the Structure pane. These panes are for viewing the names of tables, joins, and contexts. When you click a component in a pane, its corresponding graphical representation in the schema is highlighted. Graphic options, which let you customize the shape or appearance of the tables, columns, joins, and cardinalities in the Structure pane.

Arrange tables, a feature that reorganizes the tables in the Structure pane so as to produce an orderly display. Gridlines, a command that displays a grid, which you can use to align tables in the Structure pane.

Table (Column) Values, commands that display the data values associated with a particular table or column.

32. What is Join Path Problem?

A one to many join links a table, which is in turn linked by a one to many join. This type of fanning out of one to many joins can lead to a join path problem called a fan trap.

The fanning out effect of “one to many” joins can cause incorrect results to be returned when a query includes objects based on both tables.

33. How to add aggregate table in universe in real life scenario?

Decide the reports which are using high aggregates

Create Aggregate table in DB

Insert them in Universe

Join them with dimension table

34. If we have a user group and we want to give the access to the report from 1990 to 2000. and from that group we want one user to restrict to see the report from 1990 to 1995. Then what to do in BO designer, so that this can be possible( Not in report level).

This can be done using row level security in universe parameter.

35. What is shortcut join. What is its use explain with an example

A shortcut join is a join which links two table by bypassing intervening tables that exists in the universe. This is used when it is possible in certain circumstances to make SQL more efficient.

e.g. in Above example if you want get the client list and there countries you can simple join country ID from client table to country id from country table. However this would introduce a loop. So instead of using normal join you make it as shortcut join. So if query contains object from client table and country table it would you shortcut join. This would result in efficient query avoiding extra join of region.

36. What is isolated join, explain with an example.

Isolated joins means the joins which are not included in

any context. suppose you have 15 joins in ur universe. in

context A you included 7 joins and in context B 7 joins.

The rest means one join is ur isolated join.

37. What is cardinality and its significance in BO universe

Cardinality expresses the minimum and maximum number of instances of an entity B that can be associated with an instance of an entity A. The minimum and the maximum number of instances can be equal to 0, 1, or N.

Because a join represents a bi-directional relationship, it must always have two cardinalities.

There are two main methods for detecting or editing cardinalities:

• Detect Cardinalities command

• Edit Join dialog box

If you selected the Detect cardinalities in joins options in the Database tab of the Options dialog box, Designer detects and retrieves the cardinalities of the joins. If you do not use this option, you can still retrieve the cardinalities for one or all joins in the universe.

SQL QUERIES

1) Display the details of all employees
SQL>Select * from emp;

2) Display the depart information from department table
SQL>select * from dept;

3) Display the name and job for all the employees
SQL>select ename,job from emp;

4) Display the name and salary  for all the employees
SQL>select ename,sal from emp;

5) Display the employee no and totalsalary  for all the employees
SQL>select empno,ename,sal,comm, sal+nvl(comm,0) as”total  salary” from
emp

6) Display the employee name and annual salary for all employees.
SQL>select ename, 12*(sal+nvl(comm,0)) as “annual Sal” from emp

7) Display the names of all the employees who are working in depart number 10.
SQL>select emame from emp where deptno=10;

Display the names of all the employees who are working as clerks and
drawing a salary more than 3000.
SQL>select ename from emp where job=’CLERK’ and sal>3000;

9) Display the employee number and name  who are earning comm.
SQL>select empno,ename from emp where comm is not null;

10) Display the employee number and name  who do not earn any comm.
SQL>select empno,ename from emp where comm is null;

11) Display the names of employees who are working as clerks,salesman or
analyst and drawing a salary more than 3000.
SQL>select ename  from emp where job=’CLERK’ OR JOB=’SALESMAN’
OR JOB=’ANALYST’ AND SAL>3000;

12) Display the names of the employees who are working in the company for
the past 5 years;

SQL>select ename  from emp where to_char(sysdate,’YYYY’)-to_char(hiredate,’YYYY’)>=5;

13) Display the list of employees who have joined the company before
30-JUN-90 or after 31-DEC-90.
a)select ename from emp where hiredate < ’30-JUN-1990′ or hiredate >
’31-DEC-90′;

14) Display current Date.
SQL>select sysdate from dual;

15) Display the list of all users in your database(use catalog table).
SQL>select username from all_users;

16) Display the names of all tables from current user;
SQL>select tname from tab;

17) Display the name of the current user.
SQL>show user

18) Display the names of employees working in depart number 10 or 20 or 40
or employees working as
CLERKS,SALESMAN or ANALYST.
SQL>select ename from emp where deptno in(10,20,40) or job
in(‘CLERKS’,’SALESMAN’,’ANALYST’);

19) Display the names of employees whose name starts with alaphabet S.
SQL>select ename from emp where ename like ‘S%’;

20) Display the Employee names for employees whose name ends with alaphabet S.


SQL>select ename from emp where ename like ‘%S’;

21) Display the names of employees whose names have second alphabet A in
their names.
SQL>select ename from emp where ename like ‘_A%’;

22) select the names of the employee whose names is exactly five characters
in length.
SQL>select ename from emp where length(ename)=5;

23) Display the names of the employee who are not working as MANAGERS.
SQL>select ename from emp where job not in(‘MANAGER’);

24) Display the names of the employee who are not working as SALESMAN OR
CLERK OR ANALYST.
SQL>select ename from emp where job not
in(‘SALESMAN’,’CLERK’,’ANALYST’);

25) Display all rows from emp table.The system should wait after every
screen full of informaction.
SQL>set pause on

26) Display the total number of employee working in the company.
SQL>select count(*) from emp;

27) Display the total salary beiging paid to all employees.
SQL>select sum(sal) from emp;

28) Display the maximum salary from emp table.
SQL>select max(sal) from emp;

29) Display the minimum salary from emp table.
SQL>select min(sal) from emp;

30) Display the average salary from emp table.
SQL>select avg(sal) from emp;

31) Display the maximum salary being paid to CLERK.
SQL>select max(sal) from emp where job=’CLERK’;

32) Display the maximum salary being paid to depart number 20.
SQL>select max(sal) from emp where deptno=20;

33) Display the minimum salary being paid to any SALESMAN.
SQL>select min(sal) from emp where job=’SALESMAN’;

34) Display the average salary drawn by MANAGERS.
SQL>select avg(sal) from emp where job=’MANAGER’;

35) Display the total salary drawn by ANALYST working in depart number 40.
SQL>select sum(sal) from emp where job=’ANALYST’ and deptno=40;

36) Display the names of the employee in order of salary i.e the name of
the employee earning lowest salary    should salary appear first.
SQL>select ename from emp order by sal;

37) Display the names of the employee in descending order of salary.
a)select ename from emp order by sal desc;

38) Display the names of the employee in order of employee name.
a)select ename from emp order by ename;

39) Display empno,ename,deptno,sal sort the output first base on name and
within name by deptno and with in deptno by sal.
SQL>select empno,ename,deptno,sal from emp order by

40) Display the name of the employee along with their annual salary(sal*12).The name of the employee earning highest annual salary should apper first.
SQL>select ename,sal*12 from emp order by sal desc;

41) Display name,salary,hra,pf,da,total salary for each employee. The
output should be in the order of total salary,hra 15% of salary,da 10% of salary,pf 5%
salary,total salary will be(salary+hra+da)-pf.
SQL>select ename,sal,sal/100*15 as hra,sal/100*5 as pf,sal/100*10 as
da, sal+sal/100*15+sal/100*10-sal/100*5 as total from emp;

42) Display depart numbers and total number of employees working in each
department.
SQL>select deptno,count(deptno)from emp group by deptno;

43) Display the various jobs and total number of employees within each job
group.
SQL>select job,count(job)from emp group by job;

44) Display the depart numbers and total salary for each department.
SQL>select deptno,sum(sal) from emp group by deptno;

45) Display the depart numbers and max salary for each department.
SQL>select deptno,max(sal) from emp group by deptno;

46) Display the various jobs and total salary for each job
SQL>select job,sum(sal) from emp group by job;

47) Display the various jobs and total salary for each job
SQL>select job,min(sal) from emp group by job;

48) Display the depart numbers with more than three employees in each dept.
SQL>select deptno,count(deptno) from emp group by deptno having
count(*)>3;

49) Display the various jobs along with total salary for each of the jobs
where total salary is greater than 40000.
SQL>select job,sum(sal) from emp group by job having sum(sal)>40000;

50) Display the various jobs along with total number of employees in each
job.The output should contain only those  jobs with more than three employees.
SQL>select job,count(empno) from emp group by job having count(job)>3

51) Display the name of the empployee who earns highest salary.
SQL>select ename from emp where sal=(select max(sal) from emp);

52) Display the employee number and name for employee working as clerk and
earning highest salary among clerks.
SQL>select empno,ename from emp where where job=’CLERK’
and sal=(select max(sal) from emp  where job=’CLERK’);

53) Display the names of salesman who earns a salary more than the highest
salary of any clerk.
SQL>select ename,sal from emp where job=’SALESMAN’ and sal>(select
max(sal) from emp
where job=’CLERK’);

54) Display the names of clerks who earn a salary more than the lowest
salary of any salesman.
SQL>select ename from emp where job=’CLERK’ and sal>(select min(sal)
from emp
where job=’SALESMAN’);

Display the names of employees who earn a salary more than that of
Jones or that of salary grether than   that of scott.
SQL>select ename,sal from emp where sal>
(select sal from emp where ename=’JONES’)and sal> (select sal from emp
where ename=’SCOTT’);

55) Display the names of the employees who earn highest salary in their
respective departments.
SQL>select ename,sal,deptno from emp where sal in(select max(sal) from
emp group by deptno);

56) Display the names of the employees who earn highest salaries in their
respective job groups.
SQL>select ename,sal,job from emp where sal in(select max(sal) from emp
group by job)

57) Display the employee names who are working in accounting department.
SQL>select ename from emp where deptno=(select deptno from dept where
dname=’ACCOUNTING’)

58) Display the employee names who are working in Chicago.
SQL>select ename from emp where deptno=(select deptno from dept where
LOC=’CHICAGO’)

59) Display the Job groups having total salary greater than the maximum
salary for managers.
SQL>SELECT JOB,SUM(SAL) FROM EMP GROUP BY JOB HAVING SUM(SAL)>(SELECT
MAX(SAL) FROM EMP WHERE JOB=’MANAGER’);

60) Display the names of employees from department number 10 with salary
grether than that of any employee working in other department.
SQL>select ename from emp where deptno=10 and sal>any(select sal from
emp where deptno not in 10).

61) Display the names of the employees from department number 10 with
salary greater than that of all employee working in other departments.
SQL>select ename from emp where deptno=10 and sal>all(select sal from
emp where deptno not in 10).

62) Display the names of the employees in Uppercase.
SQL>select upper(ename)from emp

63) Display the names of the employees in Lowecase.
SQL>select lower(ename)from emp

64) Display the names of the employees in Propercase.
SQL>select initcap(ename)from emp;

65) Display the length of Your name using appropriate function.


SQL>select length(‘name’) from dual

66) Display the length of all the employee names.
SQL>select length(ename) from emp;

67) select name of the employee concatenate with employee number.
SQL>select ename||empno from emp;

68) User appropriate function and extract 3 characters starting from 2
characters from the following  string ‘Oracle’. i.e the out put should be ‘ac’.
SQL>select substr(‘oracle’,3,2) from dual

69) Find the First occurance of character ‘a’ from the following string i.e
‘Computer Maintenance Corporation’.
SQL>SELECT INSTR(‘Computer Maintenance Corporation’,’a’,1) FROM DUAL

70) Replace every occurance of alphabhet A with B in the string Allens(use
translate function)
SQL>select translate(‘Allens’,’A’,’B’) from dual

71) Display the informaction from emp table.Where job manager is found it
should be displayed as boos(Use replace function).
SQL>select replace(JOB,’MANAGER’,’BOSS’) FROM EMP;

72) Display empno,ename,deptno from emp table.Instead of display department
numbers display the related department name(Use decode function).
SQL>select empno,ename,decode(deptno,10,’ACCOUNTING’,20,’RESEARCH’,30,’SALES’,40,’OPRATIONS’) from emp;

73) Display your age in days.
SQL>select to_date(sysdate)-to_date(’10-sep-77′)from dual

74) Display your age in months.
SQL>select months_between(sysdate,’10-sep-77′) from dual

75) Display the current date as 15th Augest Friday Nineteen Ninety Saven.
SQL>select to_char(sysdate,’ddth Month day year’) from dual

76) Display the following output for each row from emp table.
scott has joined the company on wednesday 13th August ninten nintey.
SQL>select ENAME||’ HAS JOINED THE COMPANY ON  ‘||to_char(HIREDATE,’day
ddth Month  year’)   from EMP;

77) Find the date for nearest saturday after current date.
SQL>SELECT NEXT_DAY(SYSDATE,’SATURDAY’)FROM DUAL;

78) Display current time.
SQL>select to_char(sysdate,’hh:MM:ss’) from dual.

79) Display the date three months Before the current date.
SQL>select add_months(sysdate,3) from dual;

80) Display the common jobs from department number 10 and 20.
SQL>select job from emp where deptno=10 and job in(select job from emp
where deptno=20);

81) Display the jobs found in department 10 and 20 Eliminate duplicate jobs.
SQL>select distinct(job) from emp where deptno=10 or deptno=20
(or)
SQL>select distinct(job) from emp where deptno in(10,20);

82) Display the jobs which are unique to department 10.
SQL>select distinct(job) from emp where deptno=10

83) Display the details of those who do not have any person working under them.
SQL>select e.ename from emp,emp e where emp.mgr=e.empno group by
e.ename having count(*)=1;

84) Display the details of those employees who are in sales department and
grade is 3.

SQL>select * from emp where deptno=(select deptno from dept where
dname=’SALES’)and sal between(select losal from salgrade where grade=3)and
(select hisal from salgrade where grade=3);

85) Display those who are not managers and who are managers any one.
i)display the managers names
SQL>select distinct(m.ename) from emp e,emp m where m.empno=e.mgr;

ii)display the who are not managers
SQL>select ename from emp where ename not in(select distinct(m.ename)
from emp e,emp m where m.empno=e.mgr);

86) Display those employee whose name contains not less than 4 characters.
SQL>select ename from emp where length(ename)>4;

87) Display those department whose name start with “S” while the location
name ends with “K”.
SQL>select dname from dept where dname like ‘S%’ and loc like ‘%K’;

88) Display those employees whose manager name is JONES.
SQL>select p.ename from emp e,emp p where e.empno=p.mgr and
e.ename=’JONES’;

89) Display those employees whose salary is more than 3000 after giving 20%
increment.
SQL>select ename,sal from emp where (sal+sal*.2)>3000;

90) Display all employees while their dept names;
SQL>select ename,dname from emp,dept where emp.deptno=dept.deptno

91) Display ename who are working in sales dept.
SQL>select ename from emp where deptno=(select deptno from dept where
dname=’SALES’);

92) Display employee name,deptname,salary and comm for those sal in between
2000 to 5000 while location is chicago.
SQL>select ename,dname,sal,comm from emp,dept where sal  between 2000
and 5000
and loc=’CHICAGO’ and emp.deptno=dept.deptno;

93)Display those employees whose salary greter than his manager salary.
SQL>select p.ename from emp e,emp p where e.empno=p.mgr and p.sal>e.sal

94) Display those employees who are working in the same dept where his
manager is work.
SQL>select p.ename from emp e,emp p where e.empno=p.mgr and
p.deptno=e.deptno;

95) Display those employees who are not working under any manager.
SQL>select ename from emp where mgr is null

96) Display grade and employees name for the dept no 10 or 30 but grade is
not 4 while joined the company before 31-dec-82.
SQL>select ename,grade from emp,salgrade where sal between losal and
hisal and deptno     in(10,30) and grade<>4 and hiredate<’31-DEC-82′;

97) Update the salary of each employee by 10% increment who are not
eligiblw for commission.
SQL>update emp set sal=sal+sal*10/100 where comm is null;

98) SELECT those employee who joined the company before 31-dec-82 while
their dept location is newyork or  Chicago.
SQL>SELECT EMPNO,ENAME,HIREDATE,DNAME,LOC FROM EMP,DEPT
WHERE (EMP.DEPTNO=DEPT.DEPTNO)AND
HIREDATE <’31-DEC-82′ AND DEPT.LOC IN(‘CHICAGO’,’NEW YORK’);

99) DISPLAY EMPLOYEE NAME,JOB,DEPARTMENT,LOCATION FOR ALL WHO ARE WORKING
AS  MANAGER?
SQL>select ename,JOB,DNAME,LOCATION from emp,DEPT where mgr is not
null;

100) DISPLAY THOSE EMPLOYEES WHOSE MANAGER NAME IS JONES? –
[AND ALSO DISPLAY THEIR MANAGER NAME]?
SQL> SELECT P.ENAME FROM EMP E, EMP P WHERE E.EMPNO=P.MGR AND
E.ENAME=’JONES’;

101) Display name and salary of ford if his salary is equal to hisal of his
grade
a)select ename,sal,grade from emp,salgrade where sal between losal and
hisal
and ename =’FORD’ AND HISAL=SAL;

102) Display employee name,job,depart name ,manager name,his grade and make
out an under department wise?
SQL>SELECT E.ENAME,E.JOB,DNAME,EMP.ENAME,GRADE FROM EMP,EMP
E,SALGRADE,DEPT
WHERE EMP.SAL BETWEEN LOSAL AND HISAL AND EMP.EMPNO=E.MGR
AND EMP.DEPTNO=DEPT.DEPTNO ORDER BY DNAME

103) List out all employees name,job,salary,grade and depart name for every
one in the company  except ‘CLERK’.Sort on salary display the highest salary?
SQL>SELECT ENAME,JOB,DNAME,SAL,GRADE FROM EMP,SALGRADE,DEPT WHERE
SAL BETWEEN LOSAL AND HISAL AND EMP.DEPTNO=DEPT.DEPTNO AND JOB
NOT IN(‘CLERK’)ORDER BY SAL ASC;

104) Display the employee name,job and his manager.Display also employee who are without manager? SQL>select e.ename,e.job,eMP.ename AS Manager from emp,emp e where

emp.empno(+)=e.mgr

105) Find out the top 5 earners of company?
SQL>SELECT DISTINCT SAL FROM EMP E WHERE 5>=(SELECT COUNT(DISTINCT SAL)
FROM
EMP A WHERE A.SAL>=E.SAL)ORDER BY SAL DESC;

106) Display name of those employee who are getting the highest salary?
SQL>select ename from emp where sal=(select max(sal) from emp);

107) Display those employee whose salary is equal to average of maximum and
minimum?
SQL>select ename from emp where sal=(select max(sal)+min(sal)/2 from
emp);

108) Select count of employee in each department  where count greater than 3?
SQL>select count(*) from emp group by deptno having count(deptno)>3

109) Display dname where at least 3 are working and display only department
name?
SQL>select distinct d.dname from dept d,emp e where d.deptno=e.deptno
and 3>any
(select count(deptno) from emp group by deptno)

110) Display name of those managers name whose salary is more than average
salary of his company?
SQL>SELECT E.ENAME,EMP.ENAME FROM EMP,EMP E
WHERE EMP.EMPNO=E.MGR AND E.SAL>(SELECT AVG(SAL) FROM EMP);

111)Display those managers name whose salary is more than average salary of
his employee?
SQL>SELECT DISTINCT EMP.ENAME FROM EMP,EMP E WHERE
E.SAL <(SELECT AVG(EMP.SAL) FROM EMP
WHERE EMP.EMPNO=E.MGR GROUP BY EMP.ENAME) AND
EMP.EMPNO=E.MGR;

112) Display employee name,sal,comm and net pay for those employee
whose net pay is greter than or equal to any other employee salary of
the company?
SQL>select ename,sal,comm,sal+nvl(comm,0) as NetPay from emp
where sal+nvl(comm,0) >any (select sal from emp)

113) Display all employees names with total sal of company with each
employee name?
SQL>SELECT ENAME,(SELECT SUM(SAL)  FROM EMP) FROM EMP;

114) Find out last 5(least)earners of the company.?
SQL>SELECT DISTINCT SAL FROM EMP E WHERE
5>=(SELECT COUNT(DISTINCT SAL) FROM EMP A WHERE
A.SAL<=E.SAL)
ORDER BY SAL DESC;

115) Find out the number of employees whose salary is greater than their
manager salary?
SQL>SELECT E.ENAME FROM EMP ,EMP E WHERE EMP.EMPNO=E.MGR
AND EMP.SAL<E.SAL;

116) Display those department where no employee working?
SQL>select dname from emp,dept where emp.deptno not in(emp.deptno)

117) Display those employee whose salary is ODD value?
SQL>select * from emp where sal<0;

118) Display those employee whose salary contains alleast 3 digits?
SQL>select * from emp where length(sal)>=3;

119) Display those employee who joined in the company in the month of Dec?
SQL>select ename from emp where to_char(hiredate,’MON’)=’DEC’;

120) Display those employees whose name contains “A”?
SQL>select ename from emp where instr(ename,’A’)>0;
or
SQL>select ename from emp where ename like(‘%A%’);

121) Display those employee whose deptno is available in salary?
SQL>select emp.ename from emp, emp e where emp.sal=e.deptno;

122) Display those employee whose first 2 characters from hiredate -last 2
characters of salary?
SQL>select ename,SUBSTR(hiredate,1,2)||ENAME||substr(sal,-2,2) from emp

123) Display those employee whose 10% of salary is equal to the year of
joining?
SQL>select ename from emp where to_char(hiredate,’YY’)=sal*0.1;

124) Display those employee who are working in sales or research?
SQL>SELECT ENAME FROM EMP WHERE DEPTNO IN(SELECT DEPTNO FROM DEPT WHERE
DNAME IN(‘SALES’,’RESEARCH’));

125) Display the grade of jones?
SQL>SELECT ENAME,GRADE FROM EMP,SALGRADE
WHERE SAL BETWEEN LOSAL AND HISAL AND Ename=’JONES’;

126) Display those employees who joined the company before 15 of the month?
a)select ename from emp where to_char(hiredate,’DD’)<15;

127) Display those employee who has joined before 15th of the month.
a)select ename from emp where to_char(hiredate,’DD’)<15;

128) Delete those records where no of employees in a particular department
is less than 3.
SQL>delete from emp where deptno=(select deptno from emp
group by deptno having count(deptno)<3);

129) Display the name of the department where no employee working.
SQL> SELECT E.ENAME,E.JOB,M.ENAME,M.JOB FROM EMP E,EMP M
WHERE E.MGR=M.EMPNO

130) Display those employees who are working as manager.
SQL>SELECT M.ENAME MANAGER FROM EMP M ,EMP E
WHERE E.MGR=M.EMPNO GROUP BY M.ENAME

131) Display those employees whose grade is equal to any number of sal but
not equal to first number of sal?
SQL> SELECT ENAME,GRADE FROM EMP,SALGRADE
WHERE GRADE NOT IN(SELECT SUBSTR(SAL,0,1)FROM EMP)

132) Print the details of all the employees who are Sub-ordinate to BLAKE?
SQL>select emp.ename from emp, emp e where emp.mgr=e.empno and
e.ename=’BLAKE’;

133) Display employee name and his salary whose salary is greater than
highest average of department number?
SQL>SELECT SAL FROM EMP WHERE SAL>(SELECT MAX(AVG(SAL)) FROM EMP
GROUP BY DEPTNO);

134) Display the 10th record of emp table(without using rowid)
SQL>SELECT * FROM EMP WHERE ROWNUM<11
MINUS
SELECT * FROM EMP WHERE ROWNUM<10

135) Display the half of the ename’s in upper case and remaining lowercase?
SQL>SELECT
SUBSTR(LOWER(ENAME),1,3)||SUBSTR(UPPER(ENAME),3,LENGTH(ENAME))
FROM EMP;

136) Display the 10th record of emp table without using group by and rowid?
SQL>SELECT * FROM EMP WHERE ROWNUM<11
MINUS
SELECT * FROM EMP WHERE ROWNUM<10

Delete the 10th record of emp table.
SQL>DELETE FROM EMP WHERE EMPNO=(SELECT EMPNO FROM EMP WHERE ROWNUM<11
MINUS
SELECT EMPNO FROM EMP WHERE ROWNUM<10)

137) Create a copy of emp table;SQL>create table new_table as select * from emp where 1=2;

138) Select ename if ename exists more than once.
SQL>select ename  from emp e group by ename having count(*)>1;

139) Display all enames in reverse order?(SMITH:HTIMS).
SQL>SELECT REVERSE(ENAME)FROM EMP;

140) Display those employee whose joining of month and grade is equal.
SQL>SELECT ENAME FROM EMP WHERE SAL BETWEEN
(SELECT LOSAL FROM SALGRADE WHERE
GRADE=TO_CHAR(HIREDATE,’MM’)) AND
(SELECT HISAL FROM SALGRADE WHERE
GRADE=TO_CHAR(HIREDATE,’MM’));

141) Display those employee whose joining DATE is available in deptno.
SQL>SELECT ENAME FROM EMP WHERE TO_CHAR(HIREDATE,’DD’)=DEPTNO

142) Display those employees name as follows
A ALLEN
B BLAKE
SQL> SELECT SUBSTR(ENAME,1,1),ENAME FROM EMP;

143) List out the employees ename,sal,PF(20% OF SAL) from emp;
SQL>SELECT ENAME,SAL,SAL*.2 AS PF FROM EMP;

144) Create table emp with only one column empno;
SQL>Create table emp as select empno from emp where 1=2;

145) Add this column to emp table ename vrachar2(20).
SQL>alter table emp add(ename varchar2(20));

146) Oops I forgot give the primary key constraint.  Add in now.
SQL>alter table emp add primary key(empno);

147) Now increase the length of ename column to 30 characters.
SQL>alter table emp modify(ename varchar2(30));

148) Add salary column to emp table.
SQL>alter table emp add(sal number(10));

149) I want to give a validation saying that salary cannot be greater 10,000
(note give a name to this constraint)
SQL>alter table emp add constraint chk_001 check(sal<=10000)

150) For the time being I have decided that I will not impose this validation.My boss has agreed to pay more than 10,000.
SQL>again alter the table or drop constraint with  alter table emp drop constraint chk_001 (or)Disable the constraint by using  alter table emp modify constraint chk_001 disable;

151) My boss has changed his mind.  Now he doesn’t want to pay more than
10,000.so revoke that salary constraint.
SQL>alter table emp modify constraint chk_001 enable;

152) Add column called as mgr to your emp table;
SQL>alter table emp add(mgr number(5));

153) Oh! This column should be related to empno.  Give a command to add this
constraint.
SQL>ALTER TABLE EMP ADD CONSTRAINT MGR_DEPT FOREIGN KEY(MGR) REFERENCES
EMP(EMPNO)

154) Add deptno column to your emp table;
SQL>alter table emp add(deptno number(5));

155) This deptno column should be related to deptno column of dept table;
SQL>alter table emp add constraint dept_001 foreign key(deptno)
reference dept(deptno)
[deptno should be primary key]

156) Give the command to add the constraint.
SQL>alter table <table_name) add constraint <constraint_name>
<constraint type>

157) Create table called as newemp.  Using single command create this table
as well as get data into this table(use create table as);
SQL>create table newemp as select * from emp;
SQL>Create table called as newemp.  This table should contain only
empno,ename,dname.
SQL>create table newemp as select empno,ename,dname from emp,dept where
1=2;

158) Delete the rows of employees who are working in the company for more
than 2 years.
SQL>delete from emp where (sysdate-hiredate)/365>2;

159) Provide a commission(10% Comm Of Sal) to employees who are not earning
any commission.
SQL>select sal*0.1 from emp where comm is null

160) If any employee has commission his commission should be incremented by
10% of his salary.
SQL>update emp set comm=sal*.1 where comm is not null;

161) Display employee name and department name for each employee.
SQL>select empno,dname from emp,dept where emp.deptno=dept.deptno

162)Display employee number,name and location of the department in which he
is working.
SQL>select empno,ename,loc,dname from emp,dept where
emp.deptno=dept.deptno;

163) Display ename,dname even if there are no employees working in a
particular department(use outer join).
SQL>select ename,dname from emp,dept where emp.deptno=dept.deptno(+)

164) Display employee name and his manager name.
SQL>select p.ename,e.ename from emp e,emp p where e.empno=p.mgr;

165) Display the department name and total number of employees in each
department.
SQL>select dname,count(ename) from emp,dept where
emp.deptno=dept.deptno group by dname;

166)Display the department name along with total salary in each department.
SQL>select dname,sum(sal) from emp,dept where emp.deptno=dept.deptno
group by dname;

167) Display itemname and total sales amount for each item.
SQL>select itemname,sum(amount) from item group by itemname;

168) Write a Query To Delete The Repeted Rows from emp table;
SQL>Delete from emp where rowid not in(select min(rowid)from emp group
by ename)

169) TO DISPLAY 5 TO 7 ROWS FROM A TABLE
SQL>select ename from emp
where rowid in(select rowid from emp where rownum<=7
minus
select rowid from empi where rownum<5)

170)  DISPLAY  TOP N ROWS FROM TABLE?
SQL>SELECT * FROM
(SELECT *  FROM EMP ORDER BY ENAME DESC)
WHERE ROWNUM <10;

171) DISPLAY   TOP 3 SALARIES FROM EMP;
SQL>SELECT SAL FROM ( SELECT  * FROM EMP ORDER  BY SAL DESC )
WHERE ROWNUM <4

172) DISPLAY  9th FROM THE EMP TABLE?
SQL>SELECT ENAME FROM EMP
WHERE ROWID=(SELECT ROWID FROM EMP WHERE ROWNUM<=10
MINUS
SELECT ROWID FROM EMP WHERE ROWNUM <10)
select second max salary from emp;
select max(sal) fromemp where sal<(select  max(sal) from emp);

Course

Benefits of online training:
Online training offers remote students all the benefits of a classroom experience without the expense of travel, including:
Live instructors: Your instructor isn’t a recording; it’s a live person teaching the class remotely.
Real-time interaction: You can interact with the instructor to ask questions, request additional explanations, and even chat with fellow students during your class.

Labs and exercises: You get to use virtual servers and images to actually do the labs and exercises, just like in a real class.

Reduced cost: Because we don’t have to ship equipment or print manuals, we can pass the savings onto you and at the same time you can also save your valuable money and time by learning the subject online.
Online classes are much like our physical classes. Approximately half of the class is devoted to lecture, and the other half is devoted to hands-on labs.  Our classes are strictly student centered. We focus much on each and every student  for the betterment of the subject.

Online instructors come with practical knowledge and may be from any location across the globe. This allows students to be exposed to knowledge that can’t be learned in books and see how class concepts are applied in real business situations.