我正在尝试创建一个用户脚本,它将从表中提取用户名并将其放入变量中,但我遇到了一堆错误。这就是我所拥有的。
USE AP
GO
IF EXISTS (SELECT DB_ID('AP'))
DROP TABLE NewLogins
CREATE TABLE NewLogins
(LoginName VARCHAR(128))
INSERT INTO NewLogins
VALUES ('BBrown'), ('CChaplin'), ('DDyer'), ('EEbbers')
DECLARE NewLogins CURSOR
STATIC
FOR (SELECT LoginName, SUBSTRING(LoginName, 1, 4) FROM NewLogins)
DECLARE @LoginName VARCHAR(128), @Password VARCHAR(128)
OPEN NewLogins
FETCH NEXT FROM NewLogins
INTO @LoginName, @Password
DECLARE @DropLogin VARCHAR(200)
SET @DropLogin ='DROP LOGIN ' + @LoginName + ''
WHILE(@@FETCH_STATUS = 0)
BEGIN
IF EXISTS (SELECT DB_ID('AP'))
EXEC @DropLogin
FETCH NEXT FROM NewLogins
Declare @sqlstmt VARCHAR(200)
SET @sqlstmt='CREATE LOGIN '+@LoginName +' WITH PASSWORD ='''+ LOWER(@Password) +'9999'''
PRINT @sqlstmt
EXEC (@sqlstmt)
DECLARE @CreateUser VARCHAR(200)
SET @CreateUser ='CREATE USER '+@LoginName +'FOR LOGIN ' + @LoginName
DECLARE @AddMemberToGroup VARCHAR(200)
SET @AddMemberToGroup ='EXEC sp_addrolemember ' + @LoginName +', ' + 'PaymentEntry'
IF NOT EXISTS (SELECT * FROM sys.database_principals WHERE name = @LoginName)
BEGIN
EXEC @CreateUser
EXEC @AddMemberToGroup
END
END
CLOSE NewLogins
DEALLOCATE NewLogins
出于某种原因,我不断出现错误:
CREATE LOGIN BBrown WITH PASSWORD ='bbro9999'
Msg 15025, Level 16, State 1, Line 3
The server principal 'BBrown' already exists.
Msg 2812, Level 16, State 62, Line 45
Could not find stored procedure 'CREATE USER BBrownFOR LOGIN BBrown'.
Msg 2812, Level 16, State 62, Line 46
Could not find stored procedure 'EXEC sp_addrolemember BBrown, PaymentEntry'.
Msg 2812, Level 16, State 62, Line 28
Could not find stored procedure 'DROP LOGIN BBrown'.
答案 0 :(得分:4)
首先,您将始终从SELECT DB_ID('AP')
获取结果集。该结果集在它返回的一行/列中可能具有NULL
值,但它仍然存在,因此使用IF EXISTS
会出错。您需要检查IF DB_ID('AP') IS NOT NULL
。
接下来,您将自己打开SQL注入。您应该非常仔细地研究它,并理解为什么您使用的动态SQL是危险的。
最后,当使用带有动态SQL字符串的EXEC
时,需要将其放在大括号中,否则SQL会认为您正在尝试调用存储过程。 EXEC
函数和EXEC
命令的语法略有不同。所以,你需要EXEC(@CreateUser)