How can I change this request, so I can display the name and partitions in a r

How can I change this request, so I can add the ID of the table SPRIDEN
from now on gives me what I want:
 
1,543     A05     24     A01     24     BAC     24     BAE     24     A02     20     BAM     20
in a single line, but I would like to add the id and the name that is stored in the SPRIDEN table

 
SELECT sortest_pidm,
       max(decode(rn,1,sortest_tesc_code)) tesc_code1,
       max(decode(rn,1,score)) score1,
       max(decode(rn,2,sortest_tesc_code)) tesc_code2,
       max(decode(rn,2,score)) score2,
       max(decode(rn,3,sortest_tesc_code)) tesc_code3,
       max(decode(rn,3,score))  score3,
       max(decode(rn,4,sortest_tesc_code)) tesc_code4,
       max(decode(rn,4,score))  score4,
       max(decode(rn,5,sortest_tesc_code)) tesc_code5,
       max(decode(rn,5,score))  score5,
       max(decode(rn,6,sortest_tesc_code)) tesc_code6,
       max(decode(rn,6,score))  score6         
  FROM (select sortest_pidm,
               sortest_tesc_code,
               score, 
              row_number() over (partition by sortest_pidm order by score desc) rn
          FROM (select sortest_pidm,
                       sortest_tesc_code,
                       max(sortest_test_score) score
                  from sortest,SPRIDEN
                  where 
                  SPRIDEN_pidm =SORTEST_PIDM
                AND   sortest_tesc_code in ('A01','BAE','A02','BAM','A05','BAC')
                 and  sortest_pidm is not null  
                GROUP BY sortest_pidm, sortest_tesc_code))
                GROUP BY sortest_pidm;
                

Hello

That depends on whether spriden_pidm is unique, and you want to get the results.

Whenever you have a problem, post a small example of data (CREATE TABLE and INSERT, relevamnt columns only instructions) for all the tables and the results desired from these data.
If you can illustrate your problem using tables commonly available (such as in the diagrams of scott or HR) so you need not display the sample data; right after the results you want.
Whatever it is, explain how you get these results from these data.
Always tell what version of Oracle you are using.

Looks like you are doing something similar to the following.
Using the tables emp and dept of the scott schema, producing a line of production by Department showing the highest salary for each job, for a set given jobs:

DEPTNO DNAME          LOC           JOB_1   SAL_1 JOB_2   SAL_2 JOB_3   SAL_3
------ -------------- ------------- ------- ----- ------- ----- ------- -----
    20 RESEARCH       DALLAS        ANALYST  3000 MANAGER  2975 CLERK    1100
    10 ACCOUNTING     NEW YORK      MANAGER  2450 CLERK    1300
    30 SALES          CHICAGO       MANAGER  2850 CLERK     950

On each line, jobs are listed in order by the highest salary.
This seems to be similar to what you are doing. The roles played by the sortest_pidm, sortest_tesc_code and sortest_test_score in your table sortest are played by deptno, job and sal in the emp table. The roles played by the spriden_pidm, id and the name of your table spriden are played by deptno, dname and loc in the dept table.

Looks like you already have something like the query below, which produces a correct output, except that it does not include the dname and loc of the dept table columns.

SELECT    deptno
,       MAX (DECODE (rn, 1, job))     AS job_1
,       MAX (DECODE (rn, 1, max_sal))     AS sal_1
,       MAX (DECODE (rn, 2, job))     AS job_2
,       MAX (DECODE (rn, 2, max_sal))     AS sal_2
,       MAX (DECODE (rn, 3, job))     AS job_3
,       MAX (DECODE (rn, 3, max_sal))     AS sal_3
FROM       (
           SELECT    deptno
           ,          job
           ,          max_sal
           ,          ROW_NUMBER () OVER ( PARTITION BY  deptno
                                          ORDER BY          max_sal     DESC
                            )         AS rn
           FROM     (
                         SELECT    e.deptno
                   ,           e.job
                   ,           MAX (e.sal)     AS max_sal
                   FROM      scott.emp        e
                   ,           scott.dept   d
                   WHERE     e.deptno        = d.deptno
                   AND           e.job        IN ('ANALYST', 'CLERK', 'MANAGER')
                   GROUP BY  e.deptno
                   ,           e.job
                     )
       )
GROUP BY  deptno
;

Dept.DeptNo is unique, it won't be a dname and a loc for each deptno, so we can modify the query by replacing "deptno" with "deptno, dname, loc" throughout the query (except in the join condition, of course):

SELECT    deptno, dname, loc                    -- Changed
,       MAX (DECODE (rn, 1, job))     AS job_1
,       MAX (DECODE (rn, 1, max_sal))     AS sal_1
,       MAX (DECODE (rn, 2, job))     AS job_2
,       MAX (DECODE (rn, 2, max_sal))     AS sal_2
,       MAX (DECODE (rn, 3, job))     AS job_3
,       MAX (DECODE (rn, 3, max_sal))     AS sal_3
FROM       (
           SELECT    deptno, dname, loc          -- Changed
           ,          job
           ,          max_sal
           ,          ROW_NUMBER () OVER ( PARTITION BY  deptno      -- , dname, loc     -- Changed
                                          ORDER BY          max_sal      DESC
                            )         AS rn
           FROM     (
                         SELECT    e.deptno, d.dname, d.loc                    -- Changed
                   ,           e.job
                   ,           MAX (e.sal)     AS max_sal
                   FROM      scott.emp        e
                   ,           scott.dept   d
                   WHERE     e.deptno        = d.deptno
                   AND           e.job        IN ('ANALYST', 'CLERK', 'MANAGER')
                   GROUP BY  e.deptno, d.dname, d.loc                    -- Changed
                   ,           e.job
                     )
       )
GROUP BY  deptno, dname, loc                    -- Changed
;

In fact, you can continue to use just deptno in the analytical PARTITION BY clause. It may be slightly more efficient to just use deptno, as I did above, but it won't change the results if you use all 3, if there is only 1 danme and 1 loc by deptno.

Moreover, you don't need so many subqueries. You use the internal subquery to calculate the MAX and the outer subquery to calculate rn. Analytical functions are calculated after global fucntions so you can do both in the same auxiliary request like this:

SELECT    deptno, dname, loc
,       MAX (DECODE (rn, 1, job))     AS job_1
,       MAX (DECODE (rn, 1, max_sal))     AS sal_1
,       MAX (DECODE (rn, 2, job))     AS job_2
,       MAX (DECODE (rn, 2, max_sal))     AS sal_2
,       MAX (DECODE (rn, 3, job))     AS job_3
,       MAX (DECODE (rn, 3, max_sal))     AS sal_3
FROM       (
               SELECT    e.deptno, d.dname, d.loc
          ,       e.job
          ,       MAX (e.sal)     AS max_sal
          ,       ROW_NUMBER () OVER ( PARTITION BY  e.deptno
                                       ORDER BY       MAX (sal)     DESC
                                      )       AS rn
          FROM      scott.emp    e
          ,       scott.dept   d
          WHERE     e.deptno        = d.deptno
          AND       e.job                IN ('ANALYST', 'CLERK', 'MANAGER')
              GROUP BY  e.deptno, d.dname, d.loc
          ,       e.job
       )
GROUP BY  deptno, dname, loc
;

It will work in Oracle 8.1 or more. In Oracle 11, however, it is better to use the SELECT... Function PIVOT.

Tags: Database

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    Insert into Entry values (415,38,2006)
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    Hello, because you select tournament.tourtype in the SELECTION list, the answer is not really. If you have need of t.tourtype in the SELECT list, you might have:

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    from member m
    where exists (select 1
            from entry e inner join tournament t on (e.tourid=t.tourid)
            where m.memberid=e.memberid
                and upper(t.tourtype)='OPEN');
    

    Published by: Seanmacgc on May 7, 2009 02:09

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    - - - - - - - - - -

    If there is nothing wrong with the system, then you can go directly to any account chosen by:

    1 click on the Start button and in the search box, type netplwiz

    2 right-click on the shortcut resulting & select run as administrator.

    3 in netplwiz dialog box, check the option "users must enter a user name and password to use this computer". Select the user account Standard that you want to connect automatically by clicking on the account you want to highlight, and then click apply. Enter the password for this user account (when it exists) when you are prompted, or leave blank if there is no password.  Then, return to the dialog netplwiz new & still having that same account selected, uncheck "users must enter a user name and password to use this computer" then click OK.

    - - - - - - - -

    I do not pretend to create 2 accounts and more admin accounts that you use for day-to-day operations. It is a useful protection against the corruption of the user profiles.  No matter what profile can become corrupt and if that is the only admin account that you suddenly find you have serious problems. Operation with two spare admins [which serve just to allow tasks like facilities] protects these two accounts safe against the risk of corruption of the profile day and ensures that you will have access to the system if your normal accounts is screwed up.

  • Hello! I am a member of the creative cloud. I installed Photoshop 6 in my computer 2 years ago. I need now Adobe Creative Suite. How can I get? And how it would be a month?

    Hello! I am a member of the creative cloud. I installed Photoshop 6 in my computer 2 years ago. I need now Adobe Creative Suite. How can I get? And how it would be a month?

    Please see this document

    Update or change your plan of creative cloud

    or contact support to see what your options are:

    FAQ: How to contact Adobe for support?

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