• [^] # Re: Une solution

    Posté par . En réponse au message calcul sur plusieurs aggregats. Évalué à 0.

    Encore plus simple

    SELECT community_id, COUNT(u.mineur) AS nbmineurs, COUNT(u.nbvieux) AS nbvieux FROM
(SELECT users.*, age < 18 AS mineur, age > 60 AS vieux, community.id AS community_id FROM users
JOIN community ON community.iduser = users.id) u
GROUP BY community_id