-- PHP (or any other languaje) parts are hardcoded here!!!!
SELECT p.*, hma.howmuchattr
-- howmuchattr is needed by HAVING clause,
-- you can omit elsewhere (by surrounding SELECT or by programming languaje)
FROM products AS p
LEFT JOIN product_attributes AS pa ON pa.product_id = p.id
LEFT JOIN (
SELECT product_id, count(*) as howmuchattr
FROM product_attributes
GROUP BY product_id
) as hma on p.id = hma.product_id
WHERE
pa.attribute_id IN
(1,3) -- this cames from PHP (or any other languaje). Can be (1) for the other case
GROUP BY p.id
HAVING count(*) = howmuchattr;
siehe sqlfiddle
hier
siehe auch diese Antwort
