编写sql 触发器,执行触发器sql 报【消息 321,级别 15,状态 1,过程 t_BOS200000092Entry2_ins,第 10 行
"FInteger5" 不是可识别的表提示选项。如果它要作为表值函数的参数,请确保您的数据库兼容模式设置为 90。
】错误,求解,谢谢!
set ANSI_NULLS ON
set QUOTED_IDENTIFIER ON
go
ALTER TRIGGER [t_BOS200000092Entry2_ins]
ON [dbo].[t_BOS200000092Entry2]
for insert,update,delete
AS
BEGIN
insert into t_BOS200000085Entry2(FInteger5,FText,FBase,FBase1,FBase2,FText2,FInteger8,
FText3,FText4,FBase7,FInteger,FInteger1,FInteger2,FInteger3,FInteger4,FTime,
FTime1,FNOTE)
select * from t_BOS200000092Entry2(FInteger5,FText,FBase,FBase1,FBase2,FText2,FInteger8,FText3,
FText4,FBase7,FInteger,FInteger1,FInteger2,FInteger3,FInteger4,FTime,FTime1,
FNOTE)
FROM INSERTED
END
------解决方案--------------------
try this,
ALTER TRIGGER [t_BOS200000092Entry2_ins]
ON [dbo].[t_BOS200000092Entry2]
for insert,update,delete
AS
BEGIN
insert into t_BOS200000085Entry2(FInteger5,FText,FBase,FBase1,FBase2,FText2,FInteger8,
FText3,FText4,FBase7,FInteger,FInteger1,FInteger2,FInteger3,FInteger4,FTime,
FTime1,FNOTE)
select FInteger5,FText,FBase,FBase1,FBase2,FText2,FInteger8,FText3,
FText4,FBase7,FInteger,FInteger1,FInteger2,FInteger3,FInteger4,FTime,FTime1,
FNOTE from inserted
END