USA : +1 732 325 1626
IND : +91 800 811 4040
Mail ID : info@bigclasses.com

Pages

Online Training

The Certification to get a Successful Career in IBM Cognos


If you are interested in being the Cognos controller then you might use COG-205 exam. A technical expert whoever has the COG-205 certification have all the knowledge that's required to put in and configure the cognos 8 products. Candidates who take up this particular program could work since a effective team mate in handling cognos installation. There are numerous key concepts covered in cognos controller exam and this would come with technology, functions of all the product. If you find yourself getting ready for this certification, it truly is strongly prescribed to have some hand-on-experience of utilizing this product and you also needs to go about it in the documentation of the product.

The exam would focus on various areas similar to Architecture which could constitute to around 15%.This may require explaining regarding the third party supporting software along with the work that would be distributed among the list of controller components. Another 13% is related to conformance which in turn consists of topics like identification of all the required alternative software, controller database etc. Major part associated with Installation and configuration as to the product. Another small chunk goes to troubleshooting and tuning.

In an effort to ease the COG-205 test preparation, there are varied documentation and test preparation material available which in turn will certainly make it easier to lots within your test preparation. It can be strongly recommended to visit throughout the material inside the IBM website. The supplies is present in a downloadable format therefore you could download it at anytime plus use a submit as a way to work on it when you need it. The entire volume of questions within the exam is approximately 47 and the duration of the exam is just about sixty minutes. One will have to score at least 72% in an effort to pass the exam and also the exam is obtainable only in English.

Contact Us


#516,Annapurna Block,
AdityaEnclave, Ameerpet,
Hyderabad-500016.
India : Tel : +91-40-64600330.
Mobile : +91-888 602 3333.
USA : +1 757 905 2515.
Email: info@bigclasses.com

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;
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
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);

Concepts


The principal notion behind data warehousing is that the data stored for business analysis can most efficiently be accessed by dividing it from the data in the operational systems. A data warehouse, therefore, is a collection of data gathered from one or more data repositories to create a new, central database. For example an industry may create a data warehouse by extracting the operational data it has accumulated concerning the workers information, products they are working on, material they are using for making the particular product, output production and etc,. Data Warehousing is not just the data in the warehouse, but also the architecture and tools to collect, query, analyze and present information.
The characteristics of a data warehouse were first defined by W.H. Inmon who stated, “A data warehouse is subject-oriented, integrated, time-variant and non-volatile [data] collection in support of management decision making processes”. Let’s discuss  that definition down:
  • · Subject-oriented: all relevant data concerning a subject
  • · Integrated: all data in the warehouse must be compatible with each other regardless of type or location.
  • · Time-variant: all data contains a reference to time so that the age of each piece of data can be determined.
  • · Non-volatile: the data does not change once it has been collected.

Course


COURSE NAME                :   COGNOS ONLINE TRAINING

COURSE DURATION      :   30 HOURS

Introduction To Dataware Housing (Dwh)

  • introduction to DWH
  • Characteristics of DWH
  • OLTP VS Databases
  • Different Approaches of DWH (Kimball Vs Inmon)
  • Data mart Vs Data Warehouse
  • Fact Table Vs Dimension Table
  • Concepts of Schemas (Star schema & Snowflake schema)
  • Industry leading ETL and Reporting tools

Introduction To Cognos

  • About Cognos 8.4
  • 3 – Tier Architecture
  • Cognos Release (EP series7, Reportnet , Cognos 8.0/8,1/8.2/8.3) and their differences
  • Features of Cognos 8.4
  • Cognos Vs other OLAP tools
  • Cognos components (Modeling & Reporting)
  • Different cognos services

Cognos Connections

  • Personalize cognos connection
  • Create and manage portals, shortcuts, report views..etc
  • Working with Cognos configuration, content Store
  • How to Connect to different Data Sources
  • Security setup
  • Import and Export of Package (Deployment)
  • Running the Report Burn
  • Scheduling Reports

Framework Manager

  • About Framework Manager
  • Framework Manager, Metadata Modeling
  • Creation of Interface and Usage
  • Creating Model & Namespace (Layers)
  • Importing Metadata
  • Naming convention for objects in a project
  • Creating Query Subjects & Query Items
  • Merging Query Subjects
  • Creating Relationships & Setting Cardinalities
  • Creating Parameter Maps & Macros
  • Creating of Model Filters, Prompts & Calculations
  • Add Business Rules and Set Governors
  • Assigning package level security
  • Improving performance
  • Publishing Package

Query Studio

  • Introduction
  • About Ad-hoc Reports
  • Types of Report(List, Cross Tab, Chart Reports & Others)
  • Applying Filters, Prompts & Calculations
  • Sorting, Grouping, Sections & Aggregate Functions
  • Run and Manage reports

Report Studio

  • Type of Report Templates
  • Creating list, Cross Tab, Chart, repeater, Maps & Other Reports
  • Report templates, formatting Reports – fonts, style, header, footer, boarder…etc
  • Generating and Creating Prompt Page
  • Types of Prompts
  • Query Calculations & Layout Calculations
  • Filters and Dynamic Filters
  • Conditional Block
  • Conditional Formatting
  • Master Details
  • Cascading Prompt
  • HTML Items & Hyperlinks
  • Set Operators
  • Setting Burst Options
  • Drill through
  • Drill up &  Drill down
  • Setting Variables
  • Report Functions
  • Report Validation
  • Reports Run with Options

Analysis Studio

  • Insert Data
  • Change the number of Visible Items
  • Nest Data
  • Replace Data
  • Sort Data
  • Explore Data
  • Change the measure
  • Drill down for Details
  • Hide Items
  • Insert a calculation
  • Save a custom set
  • Define a filter
  • Compare sets of data
  • Using Dynamic Context

Event Studio

  • Creating Agents
  • Adding Event conditions
  • Task and Task execution rules
  • Scheduling Agents

Home


Cognos is a developer of Business Intelligence software developed by IBM. Cognos is designed for the Business users who have no technical Knowledge to extract corporate data, analyze it and assemble reports. Cognos is a combination of nearly three dozen software products.
Cognos was an Ottawa, Ontario-based company making Business Intelligence (BI) and performance management (PM) software. Founded in 1969, at its peak Cognos employed almost 3,500 people and served more than 23,000 customers in over 135 countries. Originally Quasar Systems Limited, it adopted the Cognos name in 1982. On January 31, 2008, Cognos was officially acquired by IBM. The Cognos name continues to be applied to IBM’s line of Business Intelligence and performance management products.
In 2005 IBM’s leading performance management software Cognos 8 Business Intelligence suite was first introduced. Cognos 8 Business Intelligence’s capabilities include reporting, analysis, dash boarding and scorecards, delivered through a web-based service-oriented architecture.
In 2007, following SAP’s acquisition of Business Objects and Oracle’s acquisition of Hyperion, IBM announced its acquisition of Cognos in November for $4.9 billion.It continued to operate as a wholly owned subsidiary (Cognos, an IBM company) until January 1, 2009, when it was absorbed into IBM’s Information Management brand within the company’s Software Group. The software is now called “Cognos Business Intelligence and Financial Performance Management” or Cognos BI and FPM.
in September 2009 IBM Cognos Express launched an integrated Business Intelligence and planning solution purpose-built to meet the needs of midsize companies. The features of the Cognos Express are:
Cognos Express Reporter (Self-service reporting and ad hoc query)
Cognos Express Advisor (Freeform analysis and visualization)
Cognos Express Xcelerator (Microsoft Excel-based planning and business analysis)
IBM Cognos also offers several Application Development Tools:
IBM Cognos PowerHouse 4GL
IBM Cognos PowerHouse Web
IBM Cognos Axiant 4GL
In October 2010, IBM released Cognos 10, which integrates Cognos with other IBM products, including SPSS Predictive Analytics and Lotus Connect. Cognos 10 comes with mobile capability, allowing users to access a complete version of Cognos from mobile devices such as tablets and smart phones. IBM’s Cognos products are used by over 23,000 companies around the world, and are consistently ranked in the “leaders” category of Gartner Magic Quadrants for Business Intelligence.

The main objective of Business Intelligence is to bring the right information at the right time. Usually, the data is pulled from several sources and transformed into accurate and consistent information which is stored in the Data Warehouse. Now Cognos is one of the largest suppliers of Business Intelligence Tools globally and the name is recognized as a leader in their field of Business Intelligence reporting.
TrainingCognos Trainer-lead Cognos Online Training is live, real-time and interactive. It is not self paced or pre-recorded, and provides many of the same benefits of in-person classroom training, With inter-active live chat and online collaboration features, information flow between students and their instructors or peers can equal or even surpass that of the classroom training experience.