Dimensional Modeling/Datamodeling
Quick Reference Guide to Dimensional Modeling
Dimensional modeling is the design concept used by many data warehouse designers to build their data warehouse. Dimensional model is the underlying data model used by many of the commercial OLAP products available today in the market. Designing a data warehouse is very different from designing an online transaction processing (OLTP) system. In contrast to an OLTP system in which the purpose is to capture high rates of data changes and additions, the purpose of a data warehouse is to organize large amounts of stable data for ease of analysis and retrieval. Because of these differing purposes, there are many considerations in data warehouse design that differ from OLTP database design. In dimensional model, all data is contained in two types of tables called Fact Table and Dimension Table.
Fact Table
Each data warehouse or data mart includes one or more fact tables. The fact table captures the data that measures the organizations business operations. A fact table might contain business sales events such as cash register transactions or the contributions and expenditures of a nonprofit organization. Fact tables usually contain large numbers of rows, sometimes in the hundreds of millions of records when they contain one or more years of history for a large organization. A key characteristic of a fact table is that it contains numerical data (facts) that can be summarized to provide information about the history of the operation of the organization. Each fact table also includes a multipart index that contains as foreign keys the primary keys of related dimension tables, which contain the attributes of the fact records. Fact tables should not contain descriptive information or any data other than the numerical measurement fields and the index fields that relate the facts to corresponding entries in the dimension tables. An example of fact table is Sales_Fact table that might contain the information like sale_amount, unit_price, discount, etc.
Dimension Table
Dimension tables contain attributes that describe fact records in the fact table. Some of these attributes provide descriptive information; others are used to specify how fact table data should be summarized to provide useful information to the analyst. Dimension tables contain hierarchies of attributes that aid in summarization. For example, a dimension containing product information would often contain a hierarchy that separates products into categories such as food, drink, and non-consumable items, with each of these categories further subdivided a number of times until the individual product is reached at the lowest level.
Dimensional modeling produces dimension tables in which each table contains fact attributes that are independent of those in other dimensions. For example, a customer dimension table contains data about customers, a product dimension table contains information about products, and a store dimension table contains information about stores. Queries use attributes in dimensions to specify a view into the fact information. For example, a query might use the product, store, and time dimensions to ask the question "What was the cost of non-consumable goods sold in the northeast region in 1999?" Subsequent queries might drill down along one or more dimensions to examine more detailed data, such as "What was the cost of kitchen products in New York City in the third quarter of 1999?" In these examples, the dimension tables are used to specify how a measure (sale_amount) in the fact table is to be summarized.
Consider an example of Sales_Fact table and the various attributes that describe this fact are Store, Product, Time and say Sales Person. In this case we will have four dimension tables, viz. Store_Dimension, Product_Dimension, Time_Dimension and Sales_Person_Dimension.
Figure 1
You may notice that all of these dimensions contain a Key field. This is called Surrogate Key. This key is substitute for a natural key in dimensions (e.g., in Sales_Person_Dimension, we have natural key as ID). In a data warehouse a surrogate key is a generalization of the natural production key and is one of the basic elements of data warehouse.
As a fact table is described by the four dimension tables described above, it will contain the Surrogate Keys of all these dimensions. This is how the Sales_Fact table will look like:
Figure 2
Now if you carefully look at the structure of above tables and how they are linked the schema will look like this:
Figure 3
You can easily tell that this looks like a STAR. Hence its known as Star Schema.
Advantages of having Star Schema
Star Schema is very easy to understand, even for non technical business managers
Star Schema provides better performance and smaller query times
Star schema is easily extensible and will handle future changes easily
Slowly Changing Dimensions
Handling changes to dimensional data across time is the most important aspect in designing a data warehouse. In dimensional modeling, there is a very rare chance that a dimension will remain static over time. For example, a customer address may change; a company may phase out old products and introduce new products. What if a customer name changes, sales person changes his region of sale or a company assigns new sales territory. How to record the history or preserve the old version of history? Here comes the concept of Slowly Changing Dimensions. The term Slowly Changing Dimension is about variation in dimensional attributes over time. The word slowly, in this context, might seem incorrect. A sales person may change his territory rapidly. But in general, when compared to measures in fact table, the changes in dimensions occur slowly.
Types of Slowly Changing Dimensions
In reference to Figure 3 above, lets say a sales person changes his region of sale. We may handle this change in several ways. These methods fall in various categories based on companys need to preserve an accurate history of dimensional changes. Ralph Kimball categorized the dimensional changes into three categories
Type One: Changes that overwrite history
Type Two: Preserve history
Type Three: Preserve a version of history
Type One (Overwrite History)
A type one change overwrites existing dimensional attribute with new information. In Sales Person Region change example, the old region name will be overwritten by the new region. Say, a sales person Rob, has territory as ASIA.
Sales_Person_Dimension
Sales_Person_Key
ID
Name
Region
...
100
203234
Rob Doe
ASIA
...
Now, if he starts looking after NorthWest Region, by implementing Type 1 dimension, the dimension table will look like:
Sales_Person_Dimension
Sales_Person_Key
ID
Name
Region
...
100
203234
Rob Doe
NorthWest
...
Advantages:
This is the easiest way to handle the Slowly Changing Dimension problem, since there is no need to keep track of the old information.
Disadvantages:
All history is lost. By applying this methodology, it is not possible to trace back in history. For example, in this case, the company would not be able to know that Christina lived in Illinois before.
Type Two (Preserve History)
A Type Two change writes a record with the new attribute information and preserves a record of the old dimensional data. Type Two changes let you preserve historical data. Implementing Type Two changes within a data warehouse might require significant analysis and development. Type Two changes accurately partition history across time more effectively than other types. However, because Type Two changes add records, they can significantly increase the database's size.
In our example, lets say we identify Region as Type Two attribute. This can be handled in this way using:
Sales_Person_Dimension
Sales_Person_Key
ID
Name
Region
...
100
203234
Rob Doe
ASIA
...
153
203234
Rob Doe
NorthWest
...
Advantages:
This allows us to accurately keep all historical information.
Disadvantages:
This will cause the size of the table to grow fast. In cases where the number of rows for the table is very high to start with, storage and performance can become a concern.
This necessarily complicates the ETL process.
Type Three (Preserve a Version of History)
You usually implement Type Three changes only if you have a limited need to preserve and accurately describe history, such as when someone gets married and you need to retain the previous name. Instead of creating a new dimensional record to hold the attribute change, a Type Three change places a value for the change in the original dimensional record. You can create multiple fields to hold distinct values for separate points in time. In the case of a region change example, you could create an OLD_REGION and NEW_REGION field and a REGION_CHANGE_EFF_DATE field to record when the change occurs. This method preserves the change. But how would you handle a second name change, or a third, and so on? The side effects of this method are increased table size and, more important, increased complexity of the queries that analyze historical values from these old fields. After more than a couple of iterations, queries become impossibly complex, and ultimately you're constrained by the maximum number of attributes allowed on a table.
This is how the table will look like in Type Three change:
Sales_Person_Dimension
Sales_Person_Key
ID
Name
Old Region
New Region
...
100
203234
Rob Doe
ASIA
NorthWest
...
Advantages:
This does not increase the size of the table, since new information is updated.
This allows us to keep some part of history.
Disadvantages:
Type 3 will not be able to keep all history where an attribute is changed more than once. For example, if Christina later moves to Texas on December 15, 2003, the California information will be lost.
Because most business requirements include tracking changes over time, data warehouse architects commonly implement Type Two changes. A data warehouse might use Type Two changes for all attributes in all tables. As an alternative, you can implement a mix of Type One and Type Two changes at an attribute level by implementing Type 2 changes for only attributes whose historical values are important when you're slicing and dicing. For example, users might not need to know an individual's previous name if a name change occurs, so a Type One change would suffice. Users might want the system to show only the person's current name. However, if the company reassigns sales territories, users might need to track who sold what, at what time, and in what territory, necessitating a Type Two change.
Although most data warehouses include Type Two changes, you need to seriously examine the business need to record historical data. Implementing Type Two changes might be necessary, but those changes will increase the database size, degrade performance, and lengthen the development time. You need to carefully evaluate using a Type Two implementation, a Type One implementation, or a hybrid implementation.
Source Systems
Typically in any organization the data is stored in various databases, usually divided up by the systems. There may be data for marketing, sales, payroll, engineering, etc. These systems might be legacy/mainframe systems or relational database systems.
Staging Area
The data coming from various source systems is first kept in a staging area. The staging area is used to clean, transform, combine, de-duplicate, household, archive, and to prepare source data for use in data warehouse. The data coming from source system is kept as it is in this area. This need not be based on relational terminology. Sometimes managers of the data are comfortable with normalized set of data. In these cases, normalized structure of the data staging storage is certainly acceptable. Also, staging area doesnt provide querying/presentation services.
Presentation Server
Once the data is in staging area, it is cleansed, transformed and then sent to Data warehouse. You may or may not have ODS before transferring data to Data Warehouse.
OLAP
The data in Data Warehouse has to be easily manipulated in order to answer the business questions from management and other users. This is accomplished by connecting the data to fast and easy-to-use tools known as Online Analytical Processing (OLAP) tools. OLAP tools can be thought of as super high-speed forklifts that have knowledge of the warehouse and the operators built into them in order to allow ordinary people off the street to jump in and quickly find products by asking English-like questions. Within the OLAP server, data is reorganized to meet the reporting and analysis requirements of the business, including:
Exception reporting
Ad-hoc analysis
Actual vs. budget reporting
Data mining (looking for trends or anomalies in the data)
In order to process business queries at high speed, answers to common questions are preprocessed in some OLAP servers, resulting in exceptional query responses at the cost of having an OLAP database that may be several times bigger than the data warehouse itself.
Data Mart
Data mart is a logical subset of complete data warehouse. It is often viewed as the restriction of data warehouse to a single business process or to a group of related business processes targeted toward a particular business group. For example an organization may have a data mart for Sales or Inventory
Tuesday, May 24, 2011
Oracle QUERIES 3
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
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
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);
Oracle QUERIES2
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 WORKINGAS 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);
Oracle QUERIES1
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;
8) 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
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