Order based on the particular values
Hello
Here is an example of data from the table I,
ID CAT_DESC OVERDUE_DAYS
1 DEPOSIT 45
1 DEPOT 48
10 FT 30
10 FT 40
10 FT 50
1 DEPOSIT 40
2 45 LOAN
2 43 LOAN
2 LOAN 40
My request here is to show records with CAT_DESC = "Pi" FIRST and then sort the another recordset based on the OVERDUE_DAYS DESC.
I tried the query below and I do a sort, showing all records in FT first with overdue_days desc, and other records also sorted by the column ID of the rest...
Select * from test_order
order by case when id = 10 then 1 else 2 end, id, overdue_days desc;
The result retrieved is
ID | CAT_DESC | OVERDUE_DAYS |
10 | FT | 50 |
10 | FT | 40 |
10 | FT | 30 |
1 | DEPOSIT | 48 |
1 | DEPOSIT | 45 |
1 | DEPOSIT | 40 |
2 | LOAN | 45 |
2 | LOAN | 43 |
2 | LOAN | 40 |
BUT THE RESULT I WANT IS
ID | CAT_DESC | OVERDUE_DAYS |
10 | FT | 50 |
10 | FT | 40 |
10 | FT | 30 |
1 | DEPOSIT | 48 |
1 | DEPOSIT | 45 |
2 | LOAN | 45 |
2 | LOAN | 43 |
1 | DEPOSIT | 40 |
2 | LOAN | 40 |
Help, please... Is this feasible?
.. Deleted...
Try
Select * from table_t
order by case when id = 10 then 1 other end null, overdue_days desc, cat_desc
Tags: Database
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