mysql - SQL Select only rows where multiple relationships exist

Question

Given a parent table 'parent'

╔═══════════╦══════════╗
║ PARENT_ID ║   NAME   ║
╠═══════════╬══════════╣
║         1 ║ bob      ║
║         2 ║ carol    ║
║         3 ║ stew     ║
╚═══════════╩══════════╝

and a many-many relationship table 'rel' between parent and a (here unspecified) property table

╔═══════════╦═══════════╗
║ PARENT_ID ║  PROP_ID  ║
╠═══════════╬═══════════╣
║         1 ║         5 ║
║         1 ║         1 ║
║         2 ║         5 ║
║         2 ║         4 ║
║         2 ║         1 ║
║         3 ║         1 ║
║         3 ║         3 ║
╚═══════════╩═══════════╝

How can I select all parents that have all of a specified set of relationships? E.g. with the sample data, how can I find all parents that have both property 5 and 1?

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edit: Same question but with requirement for an exact match: SQL Select only rows where exact multiple relationships exist

Answer 1

This is called Relational Division

SELECT  a.name
FROM    parent a
        INNER JOIN rel b
            ON a.parent_ID = b.parent_ID
WHERE   b.prop_id IN (1,5)
GROUP BY a.name
HAVING COUNT(*) = 2

• SQLFiddle Demo Link

UPDATE 1

if unique constraint was not enforce on prop_id for every parent_id, DISTINCT is needed on this case.

SELECT  a.name
FROM    parent a
        INNER JOIN rel b
            ON a.parent_ID = b.parent_ID
WHERE   b.prop_id IN (1,5)
GROUP BY a.name
HAVING COUNT(DISTINCT b.prop_id) = 2