25 Cze
2008
25 Cze
'08
11:37
daje w wyniku jeden rekord z klientem o pierwszym id a w cash suma dla wszystkich klientow
A może tak:
SELECT customers.id, customers.lastname, customers.name, customers.address, customercontacts.phone, SUM( cash.value ) FROM customers LEFT JOIN customercontacts ON ( customers.id = customercontacts.id ) JOIN cash ON ( customers.id = cash.customerid ) GROUP BY customers.id
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