这种结果是不是没有办法得到？来看看！

[cuike519]

table1 struct is

a b c d
1 2002-3-10 2002-3-20 tmp
1 2002-3-12 null tmp1


to this table

a b c d
1 2002-3-10 2002-3-20 tmp
1 2002-3-11 2002-3-20 tmp
1 2002-3-12 2002-2-20 tmp
...
1 2002-3-20 2002-2-20 tmp
1 2002-3-12 null tmp

就是怎么把每一个时间区间中的每一天去出来生成一条新的纪录如果c为null则不处理！只是添加记录变化b字段其他的字段只是复制！就可以了！

[cuike519]

最后以行写错了！
应该是：
1 2002-3-12 null tmp1

[playyuer]

一句休想
要多句!

[cuike519]

给个方法！！！

[CrazyFor]

TRY:
1,建立序数表
select top 8000 identity(int,1,1) as N into numtab from
(select top 100 id=1 from sysobjects) as a,
(select top 100 id=1 from sysobjects) as b,
(select top 100 id=1 from sysobjects) as c

示，

select a.a,a.b+N,a.c,a.d from
(select * from table where c is not null)a
left join Num b on a.c-a.b>=b.N
Union all
select * from table where c is null

[CrazyFor]

TRY:忘了你的BC字段可能不是日期型了。：）
select a.a,cast(a.b as datetime)+N,a.c,a.d from
(select * from table where c is not null)a
left join Num b on cast(a.c as datetime)-cast(a.b as datetime)>=b.N
Union all
select * from table where c is null

[cuike519]

用游标可以吗？好像这样简单些？
你的方法解释一下可以吗？

[CrazyFor]

调试正确的代码：

select top 8000 identity(int,0,1) as N into num from
(select top 100 id=1 from sysobjects) as a,
(select top 100 id=1 from sysobjects) as b,
(select top 100 id=1 from sysobjects) as c


select a.a,convert(char(10),cast(a.b as datetime)+N ,120) b,a.c,a.d from
(select * from table where c is not null)a
left join Num b on cast(a.c as datetime)-cast(a.b as datetime)>=b.N
Union all
select * from table where c is null

[CrazyFor]

游标效率低而且比较麻烦。

[cuike519]

可否解释一下！比如：第一个查询怎么和我的table1没什么关系？可以解释一下吗？
我不想把代码拷贝来就用！谢谢！

[cuike519]

大侠我要的b时间是变化的，您给的是不变的！！！帮帮忙！！！

[cuike519]

时间变化我搞定了！谢谢高手赐教！可否问一问您是怎么对SQL这么熟悉！可以指个方向！！！谢谢！