Selecting rows based on multiple references in second table

Question

Here's a representative sample of data. In the first table rows 2 and 3 are incorrect. A row in Table 1 should be active only when there exists a row with status B and there does not exist a row with status C in Table 2.

+----+--------+--+   +-----+-----+--------+
| ID | ACTIVE |  |   | ID  | REF | STATUS |
+----+--------+--+   +-----+-----+--------+
|  1 |      0 |  |   |   9 |   1 | A      |
|  2 |      1 |  |   | 100 |   2 | A      |
|  3 |      1 |  |   | 103 |   2 | B      |
|  4 |      1 |  |   | 104 |   2 | C      |
+----+--------+--+   | 111 |   3 | A      |
                     | 123 |   4 | A      |
                     | 126 |   4 | B      |
                     +-----+-----+--------+

How can I set active in rows 2 and 3 to 0 based on those rules?

Answer 1

You can use update along withe the logic that you describe:

update t1
    set active = (case when exists (select 1 from t2 where t2.ref = t1.id and t2.status = 'B') and
                            not exists (select 1 from t2 where t2.ref = t1.id and t2.status = 'C')
                       then 1 else 0
                  end);

Answer 2

If you only want to update the rows that should be zero and leave everything else alone than you can use

              UPDATE TABLE_A A
                 SET A.ACTIVE = 0
               WHERE A.A_ID IN
                                ( SELECT b.ref
                                   FROM TABLE_B B
                                  WHERE NOT EXISTS (SELECT 1
                                           FROM TABLE_B
                                          WHERE STATUS = 'B'
                                            AND B.REF = REF)
                                     OR EXISTS
                                  (SELECT 1
                                           FROM TABLE_B
                                          WHERE STATUS = 'C'
                                            AND B.REF = REF))
                                 AND a.active<>0;