Question

我想在一个insert语句中插入多个记录，那时我的触发器中出现错误

错误是

Msg 512, Level 16, State 1, Procedure SetIntFlag, Line 7 Subquery returned more than 1 value. This is not permitted when the subquery follows =, !=, <, <= , >, >= or when the subquery is used as an expression. The statement has been terminated.

我的触发器是

declare @act char(16)if update (first_nm)
set @act = (select acct_no from inserted)
update Client set Phn9_int = 1 where acct_no = @act

如果多个列插入其显示错误，则将其分配给一个变量。

Answer 1

假设这是整个触发器：

update Client set Phn9_int = 1 where acct_no in (
    select acct_no from inserted)

基本上，我的经验法则是如果你在触发器中看到任何标量变量，它可能会在多行操作方面被打破。例外情况是它使用游标，但游标通常可以避免使用。

Answer 2

将光标声明插入并使用它，代码看起来应该类似于bellow：

DECLARE @act char(16)
DECLARE my_Cursor CURSOR FAST_FORWARD FOR SELECT acct_no FROM INSERTED;

OPEN my_Cursor 
FETCH NEXT FROM my_Cursor into @act

WHILE @@FETCH_STATUS = 0 
BEGIN 
     UPDATE Client SET Phn9_int = 1 where acct_no = @act
     FETCH NEXT FROM my_Cursor into @act 
END

CLOSE my_Cursor
DEALLOCATE my_Cursor

Answer 3

不要在WHERE子句中使用local_variable作为条件。而是使用简单的JOIN条件

if update (first_nm)
update c
set c.Phn9_int = 1
FROM Client c JOIN inserted i ON c.acct_no = i.acct_no

如果在一个insert语句中插入多个记录，如何使用inserttrigger

3 个答案: