select name1 from (SELECT name1 FROM diqv) as temp
select name1 from ( SELECT name1 FROM diqv ) DERIVEDTBL
这条也错: select distinct name1 from ( SELECT name1 FROM diqv order by code ) t--服务器: 消息 1033,级别 15,状态 1,行 2 除非同时指定了 TOP,否则 ORDER BY 子句在视图、内嵌函数、派生表和子查询中无效。
太快了!select name1 from ( SELECT name1 FROM diqv ) a
这条也错: select distinct name1 from ( SELECT name1 FROM diqv order by code ) t--服务器: 消息 1033,级别 15,状态 1,行 2 除非同时指定了 TOP,否则 ORDER BY 子句在视图、内嵌函数、派生表和子查询中无效。
select name1 from ( SELECT name1 FROM diqv )from 后面的谓语错了
select distinct name1 from (SELECT top 100 percent name1 FROM diqv order by code) t
select distinct name1 from ( SELECT name1 FROM diqv order by code ) t 子查詢中不能用order by
select distinct name1 from (SELECT top 100 percent name1 FROM diqv order by code) t这样子是可以的,但是结果却没有按 code 排序
发现是按照 NAME1 拼音排的。
是distinct引起的,不信测试一下: select name1 from (SELECT top 100 percent name1 FROM diqv order by code) t肯定是按code排序的
select distinct name1 from (SELECT name1 FROM diqv ) t order by code
不是什么BUG,distinct本来就是这样的! 再玩一次: select name1 from (SELECT top 100 percent name1 FROM diqv a where not exists(select * from diqv where code<a.code and name1=a.name1) order by code) t
不是什么BUG,distinct本来就是这样的! 再玩一次: ---------下面怎么没有了?
select name1 from (SELECT top 100 percent name1 FROM diqv a where not exists(select * from diqv where code<a.code and name1=a.name1) order by code) t要这么复杂?晕啊!~
select distinct name1 from ( SELECT name1 FROM diqv order by code ) t--服务器: 消息 1033,级别 15,状态 1,行 2
除非同时指定了 TOP,否则 ORDER BY 子句在视图、内嵌函数、派生表和子查询中无效。
select distinct name1 from ( SELECT name1 FROM diqv order by code ) t--服务器: 消息 1033,级别 15,状态 1,行 2
除非同时指定了 TOP,否则 ORDER BY 子句在视图、内嵌函数、派生表和子查询中无效。
子查詢中不能用order by
select name1 from (SELECT top 100 percent name1 FROM diqv order by code) t肯定是按code排序的
order by code
再玩一次:
select name1 from
(SELECT top 100 percent name1 FROM diqv a
where not exists(select * from diqv where code<a.code and name1=a.name1)
order by code) t
再玩一次:
---------下面怎么没有了?
(SELECT top 100 percent name1 FROM diqv a
where not exists(select * from diqv where code<a.code and name1=a.name1)
order by code) t要这么复杂?晕啊!~