Question

I need help from a SQL guru here…

I have data in the following format:

Key  Value
100  A
100  B
300  A
400  B
500  A
500  B
500  C
600  B
600  W

If I want to find all keys with A or B and A and B, but not AB with anything else…how would I do this? Would I need a temp table or can I somehow join the table to itself?

EDIT: The desired output would be Key 100 since it has A & B, Key 300 since it has A, Key 400 since it has B. Not Key 500 because it also contains a C, and not key 600 because it also contains a W

Answer 1

This should work for you:

SELECT DISTINCT t1.`key`
FROM table1 t1
WHERE t1.value in('A', 'B')
GROUP BY t1.`key`
HAVING COUNT(t1.value) = (SELECT COUNT(t2.value) 
                          FROM Table1 t2 
                          WHERE t1.`key` = t2.`key`)

SQL Fiddle Demo

This returns only the keys:

• Has A and B only, and nothing else so key = 500 shouldn’t be returned. But key = 100 should be included.
• Has only A. As in key = 300.
• Has only B so key = 600 shouldn’t be returned since it contains one more value than B which is w. But key = 400 should be included.

Update: How is this working

If any key has a value IN('A', 'B') then it could also contains other values as well.

Thats why I added the HAVING clause:

HAVING COUNT(t1.value) = (SELECT COUNT(t2.value) 
                          FROM Table1 t2 
                          WHERE t1.`key` = t2.`key`)

The COUNT(t1.value) is compared to the total COUNT of the values for the same key, by using a correlated subquerySELECT COUNT(t2.value) FROM Table1 t2 WHERE t1.key = t2.key , for each value t1.value. So if the current key contains values other than 'A', 'B' then the COUNT(t1.value) won’t equal to the COUNT of all values for the same key. For example the key = 500 has a count = 3 but using:

SELECT DISTINCT t1.`key`
FROM table1 t1
WHERE t1.value in('A', 'B')
GROUP BY t1.`key`

Without the HAVING clause will include it as well, since it has either A' or ‘B’. But the HAVING clause eliminates this key since it has a count = 3 not equal to 2.