Defining STDEV as an aggregate function

Question

Is there a way to force BO XI R2 to recognize SQL function STDEV as an aggregate function?
I found a reference for version 6.5 in the following link but I can't implement the suggested solution:
http://www.forumtopics.com/busobj/viewtopic.php?p=701261&sid=8e2cfd855db2069cb84d9e021a5ba4a5
Thanks,
Paulo Gottlib

Answer 1

Rank syntax is
     rank( expression1, ... expression_n ) WITHIN GROUP ( ORDER BY expression1, ... expression_n )for example we are going to find employee salary and comm
select rank(1000, 500) WITHIN GROUP (ORDER BY sal, comm) from emp;The SQL statement above would return the rank of an employee with a salary of $1,000 and a commision of $500 from within the emp table.
Now we are going to find from your table like this
first of all run this query
select * from employee order by salary, JOB_IDyour result will be like this
RowNum ID JOB_ID SALARY
1     100 667 800
2     200 670 1250
3     300 670 1250
4     400 670 1600
5     500 671 2200
6     600 671 2750
7     700 671 2850----------------------------------------------------------------------------------------
select rank(2650,671) within group (order by salary, JOB_ID) from employeeit will be return value 6 because salary 2650 near to row 6 (Salary = 2750)
similarly
select rank(1900,670) within group (order by salary, JOB_ID) from employeeit will be return value 5 because salary 1900 near to row 6 (Salary = 2200)
select rank(800, 671) within group (order by salary, JOB_ID) from employee
this query return value 2 (reason is that)
salary = 800 exist on row number 1
and Job_ID = 671 exist on row_num 5 to 7due to salary and job_id both don't match so query will return those value where salary has min difference
Now check
select rank(1250, 671) within group (order by salary, JOB_ID) from employeesimilarly it will be return value 4
.... Thanks & Regards ....
--------Syed Adeel Ali-------
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