将记录从Char
列复制到Varchar
列后,我无法使用like
语句找到该行
Create database testDB
Go
USE [testDB]
GO
SET ANSI_NULLS ON
GO
SET QUOTED_IDENTIFIER ON
GO
SET ANSI_PADDING ON
GO
CREATE TABLE [dbo].[TestTable1]
(
[Col_char] [char](20) NULL,
[Col_nchar] [nchar](64) NULL,
[Col_varchar] [varchar](64) NULL,
[Col_nvarchar] [nvarchar](64) NULL
) ON [PRIMARY]
GO
SET ANSI_PADDING OFF
GO
insert into TestTable1 values ('Sumit1%', 'Sumit1%', 'Sumit1%', 'Sumit1%')
insert into TestTable1 values ('Sumit2*', 'Sumit2*', null, 'Sumit2*')
select
[Col_char], LEN([Col_char]),
[Col_nchar], LEN([Col_nchar]),
[Col_varchar], LEN([Col_varchar]),
[Col_nvarchar], LEN([Col_nvarchar])
from
TestTable1
这一行给了我搜索结果
select *
from TestTable1
where 'sumit1' like [Col_varchar]
现在我将*
替换为%
&将[Col_char]复制到[Col_varchar]列
update TestTable1
set [Col_varchar] = Replace([Col_char], '*', '%')
where [Col_char] like '%2%'
select * from TestTable1
select * from TestTable1 where 'sumit1' like [Col_varchar]
-- this line is not showing any results
select * from TestTable1 where 'sumit2' like [Col_varchar]
select
Len(Replace([Col_char], '*', '%')),
Len(Replace([Col_varchar], '*', '%')), *
from TestTable1
答案 0 :(得分:5)
当你有SET ANSI_PADDING ON
时,CHAR(20)
将始终为20个字符,方法是用空格填充右侧。
当您将其转换为varchar时,您仍然有20个字符,因此Col_varchar
值实际为"Sumit2% "
,因此您需要查找以Sumit2
开头的字符串,但也有最后一堆空格
如果使用
替换值UPDATE
TestTable1
SET
[Col_varchar] = RTRIM(REPLACE([Col_char],'*','%'))
WHERE
[Col_char] LIKE '%2%'
它应该适合你。
有关ANSI_PADDING https://msdn.microsoft.com/en-us/library/ms187403.aspx
的信息