Entrada del blog por 97213529 郭益銘

de 97213529 郭益銘 - martes, 9 de diciembre de 2008, 10:22
Todo el mundo

select dog.Provider_no from provider dog
where dog.Provider_no not in



(select distinct Provider.Provider_no from



(
SELECT Provider.Provider_no,p1.Item_No
FROM Provide p1,Provider
where p1.Provider_no='001' and Provider.Provider_no not in
(SELECT provide.Provider_no FROM Provider INNER JOIN provide ON Provider.Provider_no = provide.Provider_no where p1.Item_no = provide.Item_no)
)



)
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