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所以我需要插入值到表中,但如果這些值存在,我希望我的腳本不能插入任何東西。我幾乎已經解決了這個問題,但是我的插入語句最終被卡住了。SQL插入如果值不存在
DECLARE @billableitemID int;
set @billableitemid = (select billableitemID from dbo.treatment where
@billableitemid=treatment.id)
set @billableitemID = 256
CREATE TABLE #MembershipBenefitItem
(BenefitID int ,
BillableItemTypeID int ,
BillableItemID int
)
INSERT INTO #MembershipBenefitItem
(BenefitID ,
BillableItemTypeID ,
BillableItemID
)
VALUES (23450,1,@billableitemid),
(57256,1,@billableitemid)
select m.* from #MembershipBenefitItem AS M
left join membershipbenefititem AS M1 on m.billableitemid= m1.billableitemid
and m.billableitemtypeid= m1.billableitemtypeid
and m.benefitid= m1.benefitID
and [email protected] and M1.billableitemtypeID=1 and
M1.benefitID IN(23450,57256)
where M1.ID is null
INSERT INTO MembershipBenefitItem
(BenefitID,
BillableItemTypeID,
BillableItemID
)
select m.BenefitID, m.BillableItemTypeID, m.BillableItemID from
#MembershipBenefitItem AS M
left join membershipbenefititem AS M1 on m.billableitemid= m1.billableitemid
and m.billableitemtypeid= m1.billableitemtypeid
and m.benefitid= m1.benefitID
and [email protected] and M1.billableitemtypeID=1 and
M1.benefitID IN(23450,57256)
where M1.ID is null
DROP TABLE #MembershipBenefitItem
另一種方式做這樣的事情會在合併聲明。只是一個想法。通常雖然這用於如果存在更新其他插入 –
你卡在什麼? –
'Where ...並且不存在(鍵上的相關子查詢)' – xQbert