Using Group By with a join

Hello All,
I have a table MYTABLE with the structure
ID Number,
ID_FK Number,
SCORE Number
The table has the following data
1,1,5
2,1,6
3,2,4
3,2,6
I have a query which is
Select ID_FK, AVG(SCORE) FROM MYTABLE GROUP BY ID_FK
This returns
1,5.5
2,5
I want to join this so that I can lookup the ID_FK in
MYLOOKUPTABLE so that I can get the output as
1-Description, 5.5
2-Description, 5
The structure of LOOKUPTABLE is
ID NUMBER,
DESCRIPTION VARCHAR2(20)
I am not able to join with the GROUP BY expression as it says the Description from the lookup table is not a Group By expression.
Any help on this is appreciated.
Thanks in advance,
Sudhindra

I am not able to join with the GROUP BY expression as it says the Description from the lookup table is not a Group By expression.I'm sorry but I am really unable to see what your problem is: you just need to include teh description column in the GROUP BY clause.
SQL> SELECT t23.lu_id, lu.descr, avg(t23.sal) AS avg_sal
  2  FROM   lu, t23
  3  WHERE  t23.lu_id = lu.id
  4  GROUP  BY t23.lu_id, lu.descr
  5  /
     LU_ID DESCR         AVG_SAL
         1 CATFISH    27657.3333
         2 GUPPY               6
         3 SNAPPER    327690.667
SQL> Cheers, APC

Similar Messages

  • Group by with outer join

    Group by sum doesn't work with outer join. Can anyone please help me to get it right?
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    CREATE TABLE COMPLAINT
      CNO     NUMBER,
      REASON  VARCHAR2(15 BYTE),
      TOTAL   NUMBER
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       (CNO, REASON, TOTAL)
    Values
       (1, 'edge', 250);
    Insert into COMPLAINT
       (CNO, REASON, TOTAL)
    Values
       (2, 'edge', 250);
    Insert into COMPLAINT
       (CNO, REASON, TOTAL)
    Values
       (3, 'brst', 300);
    Insert into COMPLAINT
       (CNO, REASON, TOTAL)
    Values
       (4, 'crea', 400);
    COMMIT;
    CREATE TABLE SCOTT.COMPLAINTROLL
      CNO   NUMBER,
      ROLL  VARCHAR2(15 BYTE)
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       (2, 'roll22');
    Insert into COMPLAINTROLL
       (CNO, ROLL)
    Values
       (1, 'roll4');
    Insert into COMPLAINTROLL
       (CNO, ROLL)
    Values
       (1, 'roll3');
    Insert into COMPLAINTROLL
       (CNO, ROLL)
    Values
       (1, 'roll2');
    Insert into COMPLAINTROLL
       (CNO, ROLL)
    Values
       (1, 'roll1');
    COMMIT;
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    CNO     REASON     TOTAL
    1     edge     250
    2     edge     250
    3     brst     300
    4     crea     400
    select * from complaintroll
    CNO     ROLL
    1     roll1
    1     roll2
    1     roll3
    1     roll4
    2     roll22
    -- total of reason code edge is 500
    select reason,sum(total)
    from complaint c
    group by reason
    REASON     SUM(TOTAL)
    brst     300
    crea     400
    edge     500
    -- total of reason code edge after outer join is 1250
    select reason,sum(total)
    from complaint c,complaintroll cr
    where c.cno=cr.cno(+)
    group by reason
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    brst     300
    crea     400
    edge     1250
    {\code}
    Thanks for reading this post.                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                   

    The problem that you described is simple. The outer join duplicates all the rows from the parent table (complaint). If you want to sum a column from the parent table, then this sum includes all the duplicated rows.
    There are several solutions for this problem.
    A) One had been shown already by Cenutil. Instead of doing an outer join, you can do a subquery in the select clause that delivers the additional information from the detail table.
    SQL> select reason,
       sum(total),
       sum((select count(*) from complaintroll cr where c.cno=cr.cno)) cnt_rolls
    from complaint c
    group by c.reason;
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    crea                   400          0
    brst                   300          0
    edge                   500          5b) sum in two steps. First sum and count including the join criteria, then use this information to calculate the correct total sum.
    SQL> select reason, sum(stotal), sum(stotal/scount), sum(scount), sum(cnt_rolls)
      2  from (select reason, sum(total) stotal, count(*) scount, count(cr.cno) cnt_rolls
      3         from complaint c
      4         left join complaintroll cr on c.cno=cr.cno
      5         group by reason, c.cno
      6         )
      7   group by reason;
    REASON          SUM(STOTAL) SUM(STOTAL/SCOUNT) SUM(SCOUNT) SUM(CNT_ROLLS)
    crea                    400                400           1              0
    brst                    300                300           1              0
    edge                   1250                500           5              5
    sql> c) another option is to do the left join, but do the aggregation only one time for the parent table. Analytic functions are helpful for that. However since analytic fuinctions can't be used inside an aggregation function, we would again need an inline view.
    SQL> select reason, sum(case when rn = 1 then total end) sum_total, count(*), count(crcno)
      2  from (select row_number() over (partition by c.reason order by c.cno) rn,
      3                   c.*, cr.cno crcno
      4         from complaint c
      5         left join complaintroll cr on c.cno=cr.cno
      6         )
      7  group by reason;
    REASON           SUM_TOTAL   COUNT(*) COUNT(CRCNO)
    brst                   300          1            0
    crea                   400          1            0
    edge                   250          5            5
    SQL> Edited by: Sven W. on Feb 10, 2011 1:00 PM - formatting + column added to 2nd option

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    Hi all
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    0-----------     3700------     (null)     अनपेड बिल्स देणे
    15800-----14202-----(null)     Petty Cash
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    (null)------ (null)------ 1789     Petty Cash
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    thanking you

    This is one way you can do it without a group by:
    SQL> with t as
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      3      select 0 amt1, 22 amt2, null amt3, 'Furniture' tag from dual union all
      4      select 15800,14202,null, 'Petty Cash' from dual union all
      5      select null, null, 9109, 'Furniture' from dual union all
      6      select null, null, 1789, 'Petty Cash' from dual
      7    )
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      9  from
    10  (
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    12          ,sum(amt2) over (partition by tag) amt2
    13          ,sum(amt3) over (partition by tag) amt3
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    15          ,row_number() over (partition by t.tag order by rownum) rn
    16    from t
    17  )
    18  where rn = 1
    19  /
    TAG              AMT1       AMT2       AMT3
    Furniture           0         22       9109
    Petty Cash      15800      14202       1789

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