Re: [lms] zapytanie

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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