怎样获得newid()的ID号
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悬赏园豆：50
[已解决问题]
解决于 2008-07-20 08:28
<P>有这样的一个存储过程</P>
<P>CREATE PROCEDURE SelectProblem <BR>@n int,<BR>@sectionID int<BR>AS<BR>declare @i int<BR>set @i=0<BR>while @i<@n<BR>begin<BR>insert into PaperDetail (TitleID) values ('exec('select top '+cast(@n as varchar)+' [ID] from problem where sectionid='+cast(@sectionID as varchar)+' order by newid()')')<BR>set @i=@i+1<BR>end<BR>GO</P>
<P>在inset(TitleID) 是取不到 那个随机产生的 ［ID］ 就是插不进去 ，这该怎么办拉</P>
<P>恳请广大博友解答一下，小弟不胜感激！</P>
<P> </P>
问题补充：
很感谢大家的解答 !!! 但是还没解决,期待中.......
最佳答案
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感觉存储过程有问题，你可以把你的表和你要完成的操作做一下说明
就是想从problem表中随机取n个ID插入到PaperDetail 表中吧。
[code]
WITH temp_A as (
select ID,random = newID() ,sectionID from problem
where sectionID = @sectionID
),
temp_B as(
select ID,rn = ROW_NUMBER() OVER(ORDER BY random),sectionID FROM temp_A
)INSERT INTO PaperDetail (TitleID)
SELECT ID FROM temp_B
WHERE temp_B.rn <= @n
[/code]
其他回答(3)
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insert into PaperDetail (TitleID) values ('exec('select top '+cast(@n as varchar)+' [ID] from problem where sectionid='+cast(@sectionID as varchar)+' order by newid()')')
说实话这句本身有错误，太复杂了，你就直说是什么目的吧，也许有简单的实现方法。
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先查看'exec('select top '+cast(@n as varchar)+' [ID] from problem where sectionid='+cast(@sectionID as varchar)+' order by newid()')'这个句子是否执行成功，最好是declare @ID int
set @ID = (exec('select top '+cast(@n as varchar)+' [ID] from problem where sectionid='+cast(@sectionID as varchar)+' order by newid()')),然后再执行插入语句
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问题提得有水平.
