Question

在使用以下脚本时遇到问题：-

width: 80%; margin: 0 auto;

使用时：-

USE master

DECLARE @COMPANY CHAR(3)
declare @createdatabase char(50)
declare @fromdatabase char (50)
declare @sql1 char(100)
declare @logicaldatabasename Char(100)
declare @logicaldatabaselog char (100)

set @fromdatabase = 'a'
set @createdatabase = 'b'

SET @COMPANY = 'PUK'

;with q as (
SELECT [name] [logical_name]
FROM sys.[master_files]
WHERE [database_id] = DB_ID(@fromdatabase))
select @logicaldatabasename = q.[logical_name]
from q
where q.[logical_name] not like '%log%'
;with q as (
SELECT [name] [logical_name]
FROM sys.[master_files]
WHERE [database_id] = DB_ID(@fromdatabase))
select @logicaldatabaselog = q.[logical_name]
from q
where q.[logical_name] like '%log%'

select @logicaldatabasename AS LOGICALDATABASENAME, @logicaldatabaselog AS LOGICALDATABSELOG

--BACKUP DATABASE @fromdatabase
--TO DISK = '\folder\'+@fromdatabase+'.bak'
Set @sql1 = 'CREATE DATABASE '+@createdatabase
Exec (@sql1)
RESTORE DATABASE @createdatabase FROM DISK = '\flder\DANTEST.bak' 
WITH MOVE @logicaldatabasename TO 'folder\b.mdf',
MOVE @logicaldatabaselog TO '\folder\b_log.ldf',
REPLACE 
select * from sys.master_files where database_id = db_id(@createdatabase)
;
go
use b
go
;
DECLARE cCursor CURSOR LOCAL FAST_FORWARD FOR


#NAME?
SELECT
  IST.TABLE_NAME AS tblName
FROM
  nhba_a.INFORMATION_SCHEMA.TABLES IST
WHERE RIGHT(IST.TABLE_NAME,3) <> 'PUK' AND IST.TABLE_TYPE = 'BASE TABLE' and ist.TABLE_NAME like '%_%'

ORDER BY IST.TABLE_NAME

--
DECLARE @tblName VARCHAR(255)
--
DECLARE @sql  NVARCHAR(4000)
DECLARE @crlf CHAR(2)

SET @crlf = CHAR(13) + CHAR(10)

OPEN cCursor
FETCH cCursor
INTO @tblName

WHILE @@fetch_status = 0
BEGIN
if @tblname like '%_%'
SET @sql = 'DROP TABLE '+QUOTENAME(@tblName);
EXEC sp_executesql @sql;

  FETCH cCursor
  INTO @tblName
END

我得到的表中没有_，但是我搜索的表中只有_。

任何人都可以解释为什么它没有返回正确的结果以及sql在后台实际上在做什么吗？

我尝试了各种方法来得出正确的结果，但由于逻辑是合理的，所以无法找出问题所在。

Answer 1

使用LIKE运算符时，下划线_实际上是指任何一个字符。这意味着_在LIKE运算符中具有特殊含义。

如果要在带下划线的地方返回表名，则需要在where子句中转义下划线，例如...

WHERE TABLE_NAME like '%\_%' ESCAPE '\'

或者您也可以使用方括号来跳过下划线，而无需使用关键字ESCAPE

WHERE TABLE_NAME like '%[_]%'

为什么信息架构会用类似的运算符来带回格式错误的结果？

1 个答案: