E
GROUP BY family_id HAVING user_id IN(1,2)
Size: a a a
NI
А
SELECT IF(COUNT(DISTINCT family_id) = 1, 'Есть общая семья', 'Нет общей семьи') knowledge
FROM family_revision
WHERE user_id IN(1,2)
NI
NI
NI
mysql> select * from family_revision fr1 INNER JOIN family_revision fr2 ON fr1.family_id=fr2.family_id WHERE fr1.user_id=1 and fr2.user_id=2;
+---------+-----------+---------+-----------+
| user_id | family_id | user_id | family_id |
+---------+-----------+---------+-----------+
| 1 | 1 | 2 | 1 |
| 1 | 3 | 2 | 3 |
+---------+-----------+---------+-----------+
NI
NI
NI