SQL one-to-many match the one side by ALL in many side

Question

In the following one to many

CREATE TABLE source(id int, name varchar(10), PRIMARY KEY(id));
CREATE TABLE params(id int, source int, value int);

where params.source is a foreign key to source.id

INSERT INTO source values(1, 'yes');
INSERT INTO source values(2, 'no');

INSERT INTO params VALUES(1,1,1);
INSERT INTO params VALUES(2,1,2);
INSERT INTO params VALUES(3,1,3);

INSERT INTO params VALUES(4,2,1);
INSERT INTO params VALUES(5,2,3);
INSERT INTO params VALUES(6,2,4);

If i have a list of param values (say [1,2,3]), how do I find all the sources that have ALL of the values in the list (source 1, "yes") in SQL?

Thanks

Answer 1

SELECT s.*
FROM source AS s
 JOIN params AS p ON (p.source = s.id)
WHERE p.value IN (1,2,3)
GROUP BY s.id
HAVING COUNT(DISTINCT p.value) = 3;

You need the DISTINCT because your params.value is not prevented from having duplicates.

Answer 2

Edit Modified to handle case where there can be multiple occurances of the value for a given source.

Try this:

SELECT
    *
FROM
    source
WHERE
    (
        SELECT COUNT(DISTINCT value)
        FROM params
        WHERE params.source = source.id
          AND params.value IN (1, 2, 3)
    ) = 3

You can rewrite it to a GROUP BY as well:

SELECT
    source.*
FROM
    source
    INNER JOIN params ON params.source = source.id
WHERE
    params.value IN (1, 2, 3)
GROUP BY
    source.id,
    source.name
HAVING
    COUNT(DISTINCT params.value) = 3