【发布时间】:2017-03-22 13:51:53
【问题描述】:
我仍然得到错误代码
无法解决“Latin1_General_CI_AS”之间的排序规则冲突 和“SQL_Latin1_General_CP1_CI_AS”中的等于操作。
尽管我使用的是根据these 两个answers 的转换。这是我的代码:
-- ...
from #TmpResult a
left join #GroupMemberTable b
on (a.DBO_Owner_Login COLLATE SQL_Latin1_General_CP1_CI_AS) = (b.login_name COLLATE SQL_Latin1_General_CP1_CI_AS)
我已将它们都转换为:SQL_Latin1_General_CP1_CI_AS
谢谢
更新: 以下是全文摘录:
create table #TmpTableSec3 (database_name varchar(100), Database_Owner varchar(200), DBO_Owner_Login varchar(200), type varchar (100), privilege varchar(100), group_name varchar(500))
insert into #TmpTableSec3
select a.database_name, a.principal, a.DBO_Owner_Login, b.type, b.privilege, b.group_name
from #TmpResult a
left join #GroupMemberTable b
on (a.DBO_Owner_Login COLLATE SQL_Latin1_General_CP1_CI_AS) = (b.login_name COLLATE SQL_Latin1_General_CP1_CI_AS)
--a.DBO_Owner_Login = b.login_name
--fieldname COLLATE DATABASE_DEFAULT = otherfieldname COLLATE DATABASE_DEFAULT
where b.group_name is not null --and b.privilege = 'admin'
order by a.database_name
然后该表由以下内容填充:
insert into #TmpResult
select a.database_name, a.Owner, b.DBO_Owner_Login from #TmpTableSec1 a
join #TmpTableSec2 b
on a.database_name = b.database_name
set @cmd = 'select name, suser_sname(owner_sid) from master.sys.databases where name = '''+@name+''''
--select @cmd
insert #TmpTableSec1 exec (@cmd)
--select @dbowner = (select suser_sname(owner_sid) from master.sys.databases where name = @name)
set @cmd = 'use '+ @name +'
select db_name(), name, suser_sname(sid)
from sys.database_principals where name = ''dbo'''
--select ''@DBO'' = (select suser_sname(sid) from sys.database_principals where name = ''dbo'') '
INSERT #TmpTableSec2 exec (@cmd)
【问题讨论】:
-
问题可能出在代码的
.....部分 -
请发布您尝试执行的整个代码
-
@NickyvV 我发了更多,但我认为问题只与那一行有关
-
您正在向表中插入数据,错误很可能是某些列的排序规则定义与您插入的不同。如果问题出在
a.DBO_Owner_Login,您可能还想在SELECT中使用a.DBO_Owner_Login COLLATE SQL_Latin1_General_CP1_CI_AS(或适当的排序规则) -
偶然是原始表格中存储为文本的字段之一?
标签: sql-server sql-server-2012 collation