Question

我有2张桌子。如果我从table1中删除一条记录，那么第一个查询应检查它的pk是否作为表2中的外键存在，那么它不应该删除它应该的记录。

我使用了这个，但抛出了语法错误

DELETE FROM Setup.IncentivesDetail
INNER JOIN Employee.IncentivesDetail ON Setup.IncentivesDetail.IncentivesDetailID = Employee.IncentivesDetail.IncentiveDetail_ID
WHERE Setup.IncentivesDetail.IncentivesDetailID= @IncentivesDetailID
    AND Employee.IncentivesDetail.IncentiveDetail_ID= @IncentivesDetailID

更新：

基于下面的答案，我已经这样做了，这是正确的吗？

If Not Exists(Select * from Employee.IncentivesDetail where IncentivesDetail.IncentiveDetail_ID= @IncentivesDetailID)
        Begin
            Delete from Setup.IncentivesDetail
            WHERE Setup.IncentivesDetail.IncentivesDetailID= @IncentivesDetailID
        End
        Else
        Begin
            RAISERROR('Record cannot be deleted because assigned to an employee',16,1) 
            RETURN  
        End

Answer 1

也许是这样的？

DELETE FROM Setup.IncentivesDetail
WHERE Setup.IncentivesDetail.IncentivesDetailID= @IncentivesDetailID
  AND NOT EXISTS(SELECT 1 FROM Employee.IncentivesDetail WHERE IncentiveDetail_ID= @IncentivesDetailID)

但我必须承认，这个闻起来有点......你在干净吗？或者这是你正在做的事情吗？

Answer 2

您所描述的是外键约束的定义如果这些表之间已有外键，请确保标记为ON DELETE CASCADE。
如果是，则应将其删除并重新创建，ON DELETE CASCADE请参阅this link from MSDN了解详细信息。
如果您还没有外键约束，则需要创建一个：

ALTER TABLE Setup.IncentivesDetail 
ADD CONSTRAINT FK_Setup_IncentivesDetail_IncentivesDetailID FOREIGN KEY (IncentivesDetailID) 
    REFERENCES Employee.IncentivesDetail (IncentiveDetail_ID ) 
;

如果记录在另一个表中作为外来记录存在，如何不删除？

2 个答案: