7 Maj
2014
7 Maj
'14
15:17
W dniu 2014-05-07 14:57, Rafał Ramocki pisze:
select c.name <http://c.name>, lastname, city, zip, address, count(n.id <http://n.id>), n.name <http://n.name>, passwd FROM customers c LEFT JOIN nodes n ON n.ownerid = c.id <http://c.id> WHERE deleted=0 GROUP BY c.id <http://c.id>;
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