Query to get total counts in the given scenario

Question

Hi,
I am using Oracle 10g
I have a table A with following data
AgId     Trm     CD     S
1000     100010     12-JAN     A
1000     100019     20-MAR     A
1000     100019     20-JUL     D
1001     100011     25-JAN     A
1001     100011     20-FEB     D
1001     100011     23-MAR     A
1001     100012     31-JAN     A
1002     100013     14-FEB     A
1002     100013     05-APR     D
1002     100015     02-MAY     A
1003     100014     03-MAR     A
1003     100014     25-MAR     D
1004     100016     22-MAY     A
1004     100017     21-JUN     A
1004     100018     01-JUL     A
1005     100020     21-MAY     D
1005     100020     21-JUL     A
1005     100020     11-AUG     D
Here the overall status of AgId '1000' is A as he is having atleast 1 Trm active
Similarly, status of AgId '1001' is A
status of AgId '1002' is A
But,status of AgId '1003' is D as his trm is disconnected after activation(according to date column'CD')
Then, status of AgId '1004' is A
and finally status of AgId '1005' is D again as his trm deactivated, activated and deactivated at last
So, considering these criteria can any one give me a query to get total no of AgId's who are 'Active' and similarly for 'Deactive'

Answer 1

Hi,
when you put some code please enclose it between two lines starting with {noformat}{noformat}
i.e.:
{noformat}{noformat}
CREATE TABLE ... etc.
{noformat}{noformat}
Also do not forget to put correct statements. Your insert were missing semicolon at the end.
Coming back to your problem, I assum that the current status of any trm is the latest status so I have done what follows:
{code:sql}
SELECT agid
     , trm
     , cd
     , s
     , ROW_NUMBER () OVER (PARTITION BY agid, trm ORDER BY cd) rn
     , COUNT (*) OVER (PARTITION BY agid, trm) rn_tot
FROM a
ORDER BY agid, trm, cd;
Output:
      AGID        TRM CD        S         RN     RN_TOT
      1000     100010 12-JAN-12 A          1          1
      1000     100019 20-MAR-12 A          1          2
      1000     100019 20-JUL-12 D          2          2
      1001     100011 25-JAN-12 A          1          3
      1001     100011 20-FEB-12 D          2          3
      1001     100011 23-MAR-12 A          3          3
      1001     100012 31-JAN-12 A          1          1
      1002     100013 14-FEB-12 A          1          2
      1002     100013 05-APR-12 D          2          2
      1002     100015 02-MAY-12 A          1          1
      1003     100014 03-MAR-12 A          1          2
      1003     100014 25-MAR-12 D          2          2
      1004     100016 22-MAY-12 A          1          1
      1004     100017 21-JUN-12 A          1          1
      1004     100018 01-JUL-12 A          1          1
      1005     100020 21-MAY-12 D          1          3
      1005     100020 21-JUL-12 A          2          3
      1005     100020 11-AUG-12 D          3          3I have used to columns rn and rn_tot to find the latest entry for a specific agid, trm
Using the query above I have done the following:
WITH mydata AS
   SELECT agid
        , trm
        , cd
        , s
        , ROW_NUMBER () OVER (PARTITION BY agid, trm ORDER BY cd) rn
        , COUNT (*) OVER (PARTITION BY agid, trm) rn_tot
    FROM a
SELECT agid
     , CASE WHEN SUM (CASE s WHEN 'A' THEN 1 END) > 0 THEN 'A' ELSE 'D' END s
  FROM mydata
WHERE rn = rn_tot
GROUP BY agid;
Output:
      AGID S
      1000 A
      1001 A
      1002 A
      1003 D
      1004 A
      1005 D
{code}
Which is listing the agid and its status if at least there is one trm with final status 'A' for that agid.
To sum up everything you can do simply like this:
{code:sql}
WITH mydata AS
   SELECT agid
        , trm
        , cd
        , s
        , ROW_NUMBER () OVER (PARTITION BY agid, trm ORDER BY cd) rn
        , COUNT (*) OVER (PARTITION BY agid, trm) rn_tot
    FROM a
totdata AS
   SELECT agid
        , CASE WHEN SUM (CASE s WHEN 'A' THEN 1 END) > 0 THEN 'A' ELSE 'D' END s
     FROM mydata
    WHERE rn = rn_tot
    GROUP BY agid
SELECT S, COUNT(*) cnt
  FROM totdata
GROUP BY S;
Output:
S        CNT
D          2
A          4
{code}
Regards.
Al