Question

I have a table with data like this:

+---------+---------+
| user_id | item_id |
+---------+---------+
|     125 |     375 |
|     125 |       2 |
|     125 |      12 |
|     125 |     375 |
|     283 |       2 |
|     283 |      11 |
|     118 |     375 |
|     118 |      12 |
+---------+---------+

And I’m trying to create a query that will return all users that have ALL of the requested items, which in this case the requested items are:

item_id IN (375, 12) 

So this should be the result:

+---------+
| user_id |
+---------+
|     125 |
|     118 |
+---------+

Short of joining the table to itself, I’m not really sure how I would accomplish this.
One idea I thought of was to group it by: user_id, item_id then maybe put a count on it, but that will count the double record of user_id: 125, item_id: 375 as 2, putting the total count for user_id 125 at 3, which has it’s own set of problems such as returning users that have the same item twice vs. what I really need, which is a user that has both of the requested items.

Answer 1

You could use HAVING COUNT(DISTINCT item_id) = 2, but I like to be more explicit—the following approach allows for arbitrary boolean logic:

SELECT   user_id
FROM     my_table
WHERE    item_id IN (375, 12)
GROUP BY user_id
HAVING   SUM(item_id = 375) AND SUM(item_id = 12)

See it on sqlfiddle.

This works in MySQL because it does not have true boolean types; instead, the “boolean” result of the comparisons is in fact an integer—1 if true and 0 if false—therefore the summation over the group is zero if no match found and non-zero otherwise: such terms can be combined with logical operators due to MySQL’s implicit type conversion back to “boolean” types. In other RDBMS, one might have to use CASE:

HAVING   SUM(CASE WHEN item_id = 375 THEN 1 ELSE 0 END) > 0
     AND SUM(CASE WHEN item_id =  12 THEN 1 ELSE 0 END) > 0