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Posted on 2002-05-31
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Last Modified: 2010-05-02
Hello,

I have a table called Inst_kinds (Kind_code, Kind_name) and a table called Instruments (Inst_num, Kind_code, Status). The value of "Status" field (Integer) can be 1, 2, 3, 4 or 5. I want to get a report which will look like this:

         | Sta=1 | Sta=2 | Sta=3 | Sta=4 | Sta=5
   ------| ----- | ----- | ----- | ----- | -----
   Kind1 |   6   |   5   |   3   |   2   |   3
   Kind2 |   3   |   2   |   3   |   1   |   0
   Kind3 1   |   1   |   1   3   |   4   |   0
   Kind4 |   4   |   3   |   1   |   3   |   1
   Kind5 |   6   |   2   |   1   |   2   |   4
   ..... |  ...  |  ...  |  ...  |  ...  |  ...
   KindN |   X   |   Y   |   Z   |   P   |   Q

The data in the cells is:
Count(Inst_num) Where (Kind_code = 1) And (Stutus = 1)
Count(Inst_num) Where (Kind_code = 1) And (Stutus = 2)
Count(Inst_num) Where (Kind_code = 1) And (Stutus = 3)
Count(Inst_num) Where (Kind_code = 1) And (Stutus = 4)
Count(Inst_num) Where (Kind_code = 1) And (Stutus = 5)
Count(Inst_num) Where (Kind_code = 2) And (Stutus = 1)
Count(Inst_num) Where (Kind_code = 2) And (Stutus = 2)
etc, etc, etc, etc, etc, etc, etc, etc......

HOW CAN I DO THIS ???
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Question by:slavikn
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4 Comments
 
LVL 1

Author Comment

by:slavikn
ID: 7046725
P.S.  The table was written using a Fixed font.
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LVL 4

Expert Comment

by:gencross
ID: 7046762
You can use GROUP BY.  Here is an example of the SQL you will want to use the return the records.

I'm writing this off of the top of my head so you may need to modify the SQL some, but you get the general idea.

sSQL = "select count(Inst_num) AS RecCount,Status,Kind_name,kind_code from Instruments INNER JOIN tbl on Instruments.Kind_Code = Inst_kinds.Kind_code GROUP BY Kind_name,kind_code,Status"

Hope this helps.
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LVL 4

Accepted Solution

by:
gencross earned 50 total points
ID: 7046765
Oops!  Try this SQL...

sSQL = "select count(Inst_num) AS RecCount,Status,Kind_name,kind_code from Instruments INNER JOIN Inst_kinds on Instruments.Kind_Code = Inst_kinds.Kind_code GROUP BY Kind_name,kind_code,Status"
0
 
LVL 1

Author Comment

by:slavikn
ID: 7048520
The answer was almost correct. Anyway, I understood the idea.
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