我有以下查詢的問題:問題的別名加入statiment
SELECT
g_contac.contid, g_contac.name, g_contac.email, f_sync.foreign_key,
(
SELECT COUNT(g_cpers.cpersid)
FROM g_cpers
WHERE g_cpers.contid = g_contac.contid
) AS employee_count
FROM f_sync
FULL OUTER JOIN g_contac ON
(
g_contac.contid = f_sync.external_id AND
model = case when f_sync.employee_count = 0 then 'PRIVATE' else 'COMPANY' end
)
WHERE model = 'COMPANY' or model = 'PRIVATE' OR model IS null
當我執行它,我得到錯誤:
無效的列名稱employee_count「。
如何解決這個問題?
在聯接語句中定義的模型上的where子句不會導致選擇所有行嗎? – 2011-06-02 08:04:07
你在f_sync表中有employee_count字段嗎?如果不是,那麼錯誤是正確的。 – Arvo 2011-06-02 08:13:54
@Arvo:我沒有。我從內部select中選擇 – Dusan 2011-06-02 08:15:21