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SQL Help

All Pls help me writing SQL for the following scenario

Table 1
COL1

Table 2
COL1
COL2 ----> FK to TAB1.COL1 (1 to many)
COL3
COL4
COL5
COL6

Table 3 Is a cross reference table (For a given TAB2.COL1 there may be many TAB4.COL1 values)
COL1 -------> Foreign key to TAB2.COL1
COL2 -------> Foreign Key to TAB4.COL1

Table 4
COL1
COL2
COL3
COL4
COL5
COL6

Now the requirement is for a given TAB1.COL1 I need to find out the count of TAB2.COL3 <> TAB4.COL3
TAB2.COL4 <> TAB4.COL4
TAB2.COL5 <> TAB4.COL5
TAB2.COL6 <> TAB4.COL6

In One query The output should be something like
COUNT(COL3) = 1
COUNT(COL4) = 1
COUNT(COL5) = 1
COUNT(COL6) = 1
Mar 22 '08 #1
2 1355
sakumar9
127 Expert 100+
One possibility:

Expand|Select|Wrap|Line Numbers
  1. Your four individual queries:
  2. (select count(tab2.col3) a from tab2, tab4 where tab2.col3 <> tab4.col3)
  3. (select count(tab2.col4) b from tab2, tab4 where tab2.col4 <> tab4.col4)
  4. (select count(tab2.col5) c from tab2, tab4 where tab2.col5 <> tab4.col5)
  5. (select count(tab2.col6) d from tab2, tab4 where tab2.col6 <> tab4.col6)
  6.  
  7.  
  8. Combined query:
  9.  
  10. WITH
  11.     a (count1) AS
  12.         (select count(tab2.col3) a from tab2, tab4 where tab2.col3 <> tab4.col3
  13.         ),
  14.     b (count2) AS
  15.         (select count(tab2.col4) b from tab2, tab4 where tab2.col4 <> tab4.col4
  16.         ),
  17.     c (count3) AS
  18.         (select count(tab2.col5) c from tab2, tab4 where tab2.col5 <> tab4.col5
  19.         ),
  20.     d (count4) AS
  21.         (select count(tab2.col6) d from tab2, tab4 where tab2.col6 <> tab4.col6
  22.         )
  23.  SELECT a.count1, b.count2, c.count3, d.count4 FROM a,b,c,d;
  24.  
  25. Results:
  26. COUNT1      COUNT2      COUNT3      COUNT4
  27. ----------- ----------- ----------- -----------
  28.           0           0           0           0
  29.  
  30.   1 record(s) selected.
  31.  
Is this what you were looking for?

Regards
-- Sanjay
Mar 24 '08 #2
Sanjay,

Thank you so much this what exactly I am looking for

Sen
Mar 25 '08 #3

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