Comments on: Making up records in SQL Queries – Table Functions and 10g Model clause

By: drkalucard

drkalucard — Tue, 05 Sep 2006 08:42:50 +0000

what do you think about this?
select Job, DECODE(DEPTNO,10,sum(SAL), NULL) Dept 10,
DECODE(DEPTNO,20,sum(SAL), NULL) Dept 20,
DECODE(DEPTNO,30,sum(SAL), NULL) Dept 10,
sum(SAL) Total
from EMP
group by Job;

By: Mike

Mike — Thu, 13 Apr 2006 20:51:00 +0000

I am trying to create a view that shows the total of overdue people by, less than 30 days, 31 to 59days, and over 60 days overdue.
What is wrong with my code.

CREATE OR REPLACE VIEW csu_status_date_vu
(category, less than 30 days, 31-59 days, 60 days and greater)

AS SELECT ass_attribute9, effective_start_date

FROM PER_ALL_ASSIGNMENTS_F

WHERE (SELECT SUM (CASE
WHEN effective_start_date > TRUNC (SYSDATE – 30)
THEN 1
ELSE 0
END) “less than 30 days”,
SUM (CASE
WHEN effective_start_date = TRUNC (SYSDATE – 59)
THEN 1
ELSE 0
END) “31 -59 days”,
SUM (CASE
WHEN effective_start_date

By: Tony Andrews

Tony Andrews — Thu, 15 Sep 2005 13:12:45 +0000

Re the pivot query, isn’t this a whole lot simpler? :-

SQL> select job
2 , sum(case when deptno=10 then sal else 0 end) dept10
3 , sum(case when deptno=20 then sal else 0 end) dept20
4 , sum(case when deptno=30 then sal else 0 end) dept30
5 , sum(sal) total
6 from emp
7 group by job;

JOB DEPT10 DEPT20 DEPT30 TOTAL
——— ———- ———- ———- ———-
ANALYST 0 6000 0 6000
CLERK 1300 1900 950 4150
MANAGER 2450 2975 2850 8275
PRESIDENT 5000 0 0 5000
SALESMAN 0 0 5600 5600

By: Lucas

Lucas — Thu, 16 Dec 2004 15:16:44 +0000

OK, using the SQL MODEL clause (only in Oracle 10g I am afraid) I found the near perfect solution. Its result looks as follows:
JOB DEPT10 DEPT20 DEPT30 GRAND_TOTAL --------- ---------- ---------- ---------- ----------- CLERK 1300 1900 950 4150 SALESMAN 5600 5600 PRESIDENT 5000 5000 MANAGER 2450 2975 2850 8275 ANALYST 6000 6000

The query itself:
select job , dept10 , dept20 , dept30 , grand_total from (select distinct deptno, job, sum(sal) OVER ( PARTITION BY deptno, job) sumsal , sum(sal) over( partition by job) sumjob from emp ) model return updated rows partition by (job) dimension by (deptno) measures (sumsal, lpad(' ',10) dept10, lpad(' ',10) dept20, lpad(' ',10) dept30 , sumjob grand_total) rules upsert ( dept10 [0] = sumsal [10] , dept20 [0] = sumsal [20] , dept30 [0] = sumsal [30] , grand_total [0] = max(grand_total) [ANY] )

For more insight and an explanation, see this post Pivoting in SQL using the 10g Model Clause

By: Lucas

Lucas — Wed, 08 Dec 2004 12:19:52 +0000

OK, I missed the proper pivot. By replacing job and deptno in the above query we get the following result:

JOB       SAL_SUM1             SAL_SUM2             SAL_SUM3
--------- -------------------- -------------------- -------------
ANALYST   10:                  20: 6000             30:
CLERK     10: 1300             20: 1900             30: 950
MANAGER   10: 2450             20: 2975             30: 2850
PRESIDENT 10: 5000             20:                  30:
SALESMAN  10:                  20:                  30: 5600

The query now looks as follows:

select job
,      substr(sal_sum1, 1, 20) sal_sum1
,      substr(sal_sum2, 1, 20) sal_sum2
,      substr(sal_sum3, 1, 20) sal_sum3
,      substr(sal_sum4, 1, 20) sal_sum4
,      substr(sal_sum5, 1, 20) sal_sum5
,      substr(sal_sum6, 1, 20) sal_sum6
from ( select job
       ,      deptno||': '||sal_sum sal_sum1
       ,      lead(deptno,1) over (partition by job order by deptno)
              ||': '||lead(sal_sum,1) over (partition by job order by deptno) sal_sum2
       ,      lead(deptno,2) over (partition by job order by deptno)
              ||': '||lead(sal_sum,2) over (partition by job order by deptno) sal_sum3
       ,      lead(deptno,3) over (partition by job order by deptno)
              ||': '||lead(sal_sum,3) over (partition by job order by deptno) sal_sum4
       ,      lead(deptno,4) over (partition by job order by deptno)
              ||': '||lead(sal_sum,4) over (partition by job order by deptno) sal_sum5
       ,      lead(deptno,5) over (partition by job order by deptno)
              ||': '||lead(sal_sum,5) over (partition by job order by deptno) sal_sum6
       ,      row_number() over (partition by job order by deptno) rnk
       from ( select distinct
                     job
              ,      deptnos.deptno
              ,      sum(sal) over (partition by emp.deptno, job) sal_sum
              from   emp partition by (job)
                     right outer join
                     (select distinct deptno from emp) deptnos
                     on (emp.deptno = deptnos.deptno)
              order
              by     job
              ,      deptno
            )
     )
where rnk = 1

By: Lucas

Lucas — Wed, 08 Dec 2004 12:16:09 +0000

Maybe a better, more generic solution with less hard-coding and less coding in general - though only suitable for Oracle 10g because of the partion by clause in the outer join expression - would render the following result:

 DEPTNO SAL_SUM1             SAL_SUM2             SAL_SUM3             SAL_SUM4             SAL_SUM5             SAL_SUM6
------- -------------------- -------------------- -------------------- -------------------- -----
     10 ANALYST:             CLERK: 1300          MANAGER: 2450        PRESIDENT: 5000      SALESMAN:            :
     20 ANALYST: 6000        CLERK: 1900          MANAGER: 2975        PRESIDENT:           SALESMAN:            :
     30 ANALYST:             CLERK: 950           MANAGER: 2850        PRESIDENT:           SALESMAN: 5600       :

The code used to produce this result:

select deptno
,      substr(sal_sum1, 1, 20) sal_sum1
,      substr(sal_sum2, 1, 20) sal_sum2
,      substr(sal_sum3, 1, 20) sal_sum3
,      substr(sal_sum4, 1, 20) sal_sum4
,      substr(sal_sum5, 1, 20) sal_sum5
,      substr(sal_sum6, 1, 20) sal_sum6
from ( select deptno
       ,      job||': '||sal_sum sal_sum1
       ,      lead(job,1) over (partition by deptno order by job)
              ||': '||lead(sal_sum,1) over (partition by deptno order by job) sal_sum2
       ,      lead(job,2) over (partition by deptno order by job)
              ||': '||lead(sal_sum,2) over (partition by deptno order by job) sal_sum3
       ,      lead(job,3) over (partition by deptno order by job)
              ||': '||lead(sal_sum,3) over (partition by deptno order by job) sal_sum4
       ,      lead(job,4) over (partition by deptno order by job)
              ||': '||lead(sal_sum,4) over (partition by deptno order by job) sal_sum5
       ,      lead(job,5) over (partition by deptno order by job)
              ||': '||lead(sal_sum,5) over (partition by deptno order by job) sal_sum6
       ,      row_number() over (partition by deptno order by job) rnk
       from ( select distinct
                     deptno
              ,      jobs.job
              ,      sum(sal) over (partition by emp.job, deptno) sal_sum
              from   emp partition by (deptno)
                     right outer join
                     (select distinct job from emp) jobs
                     on (emp.job = jobs.job)
              order
              by     deptno
              ,      job
            )
     )
where rnk = 1

By: Lucas Jellema

Lucas Jellema — Wed, 24 Nov 2004 08:50:37 +0000

I am not sure this is the most elegant solution, but it most certainly gives the matrix report you asked for:
JOB DEPT10 DEPT20 DEPT30 TOTAL -------- ---------- ---------- ---------- ---------- Clerks 1430 2090 1045 4565 Managers 0 2975 2850 5825 Salesmen 0 0 3100 3100
select case dn.n when 1 then 'Clerks' when 2 then 'Managers' when 3 then 'Salesmen' else null end job , max( case when rn=1 and dn.n = 1 then clerks when rn=1 and dn.n = 2 then managers when rn=1 and dn.n = 3 then salesmen else null end) dept10 , max( case when rn=2 and dn.n = 1 then clerks when rn=2 and dn.n = 2 then managers when rn=2 and dn.n = 3 then salesmen else null end ) dept20 , max(case when rn=3 and dn.n = 1 then clerks when rn=3 and dn.n = 2 then managers when rn=3 and dn.n = 3 then salesmen else null end) dept30 , sum(case when dn.n = 1 then clerks when dn.n = 2 then managers when dn.n = 3 then salesmen else 0 end) total from (select deptno , row_number() over (order by deptno) rn , sum( case job when 'CLERK' then sal else 0 end ) clerks , sum( case job when 'MANAGER' then sal else 0 end ) Managers , sum( case job when 'SALESMAN' then sal else 0 end ) salesmen from emp group by deptno ) , (select rownum n from emp where rownum<4) dn group by case dn.n when 1 then 'Clerks' when 2 then 'Managers' when 3 then 'Salesmen' else null end

By: Jason Grundy

Jason Grundy — Tue, 23 Nov 2004 23:27:32 +0000

There are some good examples here but don’t forget that in many circumstances an alternative approach
for magicing rows into existence is to deliberately make a cartesian join to this type of in-line view:

select rownum
from {large table}
where rownum < {rows that you want}

By: sheetal

sheetal — Tue, 23 Nov 2004 15:57:04 +0000

how do i create a matrix query to display job, salary for that job based on dept no., and total salary for that job fro all departments,
job dept10 dept20 dept30 total
—- —— —— —— —–
clerk 1300 1900 950 4150

using grouping functions

By: sheetal

sheetal — Tue, 23 Nov 2004 15:56:41 +0000