OBIEE 11.1.1.6.0 RCU Installation

OBIEE 11.1.1.6.0 RCU Installation

BI 11g is now standardized with Oracle fusion middleware and hence leverage database for metadata storage just like any other middleware product. So here we are going to install BI11g metadata schema on oracle database you just installed and configured. Same process can be leveraged to install metadata schema on other supported databases as well.
If you have not done already, please download BI11g installation files and RCU installation files. Please make sure all the zipped files have downloaded correctly and unzipped into a single folder.
We can download the software from below Download page:
http://www.oracle.com/technetwork/middleware/bi-enterprise-edition/downloads/bus-intelligence-11g-165436.html


Now we will start the installation of metadata schema using RCU.

From unzipped folder for RCU, run rcu.bat file to start the utility: rcu.bat file is located under rcuHome\BIN directory.

RCU utility will launch the GUI based wizard to guide you through the steps. Click Next to continue.

Select create from the below screen and click Next

 

In the below step Make sure database type is Oracle Database and enter following information to continue. 
For e.g:
Host Name: localhost 
Port: 1521 
Service Name: XE
Username: sys
Role: sysdba
Password:  manager

 

since we are using oracle database, RCU will show a warning message that we can just ignore and can continue the installation.

 

Click OK to continue once RCU validates and initializes database configuration.

 

In the next step it will ask for the prefix to create the metadata schemas. By default it is showing 'DEV'. If we want to change the prefix we can  change it.
From the below window select appropriate schemas which we need to install.
I have selected only two which is showing in the screenshot.

Click 'Next' from the above screen to create schema and the click 'OK' once RCU shows schema creation prerequisites check completed.

IMPORTANT QUESTION IN ORACLE ASKED BY INTERVIEWER

1)select MAX salary .

select  MAX(sal) from emp;

select MAX(salary) from employees;

SELECT LEVEL,MAX(sal) FROM emp WHERE LEVEL=1 CONNECT BY PRIOR sal > sal GROUP BY LEVEL;

2)select 2nd highest salary.

select  MAX(sal) from emp where sal < (select  MAX(sal) from emp)

select  MAX(salary) from employees where salary < (select  MAX(salary) from employees)

SELECT LEVEL, MAX(sal) FROM emp WHERE LEVEL=2 CONNECT BY PRIOR sal > sal GROUP BY LEVEL;

3)select 3rd,4th.......etc  highest salary

SELECT MIN(salary) FROM (SELECT DISTINCT salary FROM employees ORDER BY salary desc ) WHERE rownum <= 3 ORDER BY salary ;

SELECT MIN(sal) FROM (SELECT DISTINCT sal FROM emp ORDER BY sal desc ) WHERE rownum <= 3 ORDER BY sal ;

SELECT LEVEL, MAX(sal) FROM emp WHERE LEVEL=4 CONNECT BY PRIOR sal > sal GROUP BY LEVEL;

4)select MIN salary

select MIN(sal) from emp

5)select 2nd MIN salary

select MIN(sal) from emp where sal > (select  MIN(sal) from emp)

6)TOP 5 EARNER OF COMPANY

SELECT LEVEL,MAX(SAL)  FROM EMP WHERE LEVEL < 6  CONNECT BY PRIOR SAL > SAL GROUP BY LEVEL ORDER BY 1

7)LAST 5 EARNER OF COMPANY

SELECT  LEVEL RANK,MIN(SAL) FROM EMP WHERE LEVEL < 6 CONNECT BY PRIOR SAL < SAL GROUP BY LEVEL ORDER BY 1

SELECT ENAME,SAL SALARY  FROM EMP  WHERE SAL>=(SELECT (MAX(SAL)+MIN(SAL))/2 FROM EMP )

SELECT MAX(SAL),MIN(SAL)  FROM EMP