求sql：按相同日期分组并显示组号

[吃饱饱吖]

如题：

create table #t(s_time datetime,na varchar(20))

insert #t
select '2013-10-1','aa'
union all
select '2013-10-1','bb'
union all
select '2013-10-1','cc'
union all
select '2013-10-15','dd'
union all
select '2013-10-15','ee'
union all
select '2013-10-20','ff'

select * from #t

/*
	希望得到的结果

	group_no	s_time	        na
	1		2013-10-01 	aa
	1		2013-10-01 	bb
	1		2013-10-01 	cc
	2		2013-10-15 	dd
	2		2013-10-15 	ee
	3		2013-10-20 	ff

*/

[-Tracy-McGrady-]

create table #t(s_time datetime,na varchar(20))
 
insert #t
select '2013-10-1','aa'
union all
select '2013-10-1','bb'
union all
select '2013-10-1','cc'
union all
select '2013-10-15','dd'
union all
select '2013-10-15','ee'
union all
select '2013-10-20','ff'
 
select DENSE_RANK()over(order by s_time) as group_no,* from #t

drop table #t

--
group_no             s_time                  na
-------------------- ----------------------- --------------------
1                    2013-10-01 00:00:00.000 aa
1                    2013-10-01 00:00:00.000 bb
1                    2013-10-01 00:00:00.000 cc
2                    2013-10-15 00:00:00.000 dd
2                    2013-10-15 00:00:00.000 ee
3                    2013-10-20 00:00:00.000 ff

(6 行受影响)

[-Tracy-McGrady-]

create table #t(s_time datetime,na varchar(20))
 
insert #t
select '2013-10-1','aa'
union all
select '2013-10-1','bb'
union all
select '2013-10-1','cc'
union all
select '2013-10-15','dd'
union all
select '2013-10-15','ee'
union all
select '2013-10-20','ff'
 
select DENSE_RANK()over(order by s_time) as group_no
,convert(char(10),s_time,120) as s_time,na 
from #t

drop table #t

--结果
group_no             s_time     na
-------------------- ---------- --------------------
1                    2013-10-01 aa
1                    2013-10-01 bb
1                    2013-10-01 cc
2                    2013-10-15 dd
2                    2013-10-15 ee
3                    2013-10-20 ff

(6 行受影响)

[chen357313771]

create table #t(s_time datetime,na varchar(20))

insert #t
select '2013-10-1','aa'
union all
select '2013-10-1','bb'
union all
select '2013-10-1','cc'
union all
select '2013-10-15','dd'
union all
select '2013-10-15','ee'
union all
select '2013-10-20','ff'

SELECT DENSE_RANK() OVER(ORDER BY s_time) AS group_no
	,CAST(s_time AS DATE) AS Date
	,na
FROM #t

/*
1	2013-10-01	aa
1	2013-10-01	bb
1	2013-10-01	cc
2	2013-10-15	dd
2	2013-10-15	ee
3	2013-10-20	ff

*/

[發糞塗牆]

 SELECT b.group_no ,
        a.*
 FROM   #t a
        INNER JOIN ( SELECT s_time ,
                            ROW_NUMBER() OVER ( ORDER BY s_time ) group_no
                     FROM   ( SELECT DISTINCT
                                        s_time
                              FROM      #t
                            ) b
                   ) b ON a.s_time = b.s_time

[Andy__Huang]

create table #t(s_time datetime,na varchar(20))
insert #t
select '2013-10-1','aa'
union all
select '2013-10-1','bb'
union all
select '2013-10-1','cc'
union all
select '2013-10-15','dd'
union all
select '2013-10-15','ee'
union all
select '2013-10-20','ff'

select group_no=(select count(distinct s_time) from #t b where b.s_time<=a.s_time),s_time,na
from #t a


/*
group_no	s_time	na
1	2013-10-01 00:00:00.000	aa
1	2013-10-01 00:00:00.000	bb
1	2013-10-01 00:00:00.000	cc
2	2013-10-15 00:00:00.000	dd
2	2013-10-15 00:00:00.000	ee
3	2013-10-20 00:00:00.000	ff

*/

[LongRui888]

通过dense_rank函数，进行密集的排名，就可以实现的：

if object_id('tb') is not null
   drop table tb
go
create table tb(s_time datetime,na varchar(20))
 
insert tb
select '2013-10-1','aa'
union all
select '2013-10-1','bb'
union all
select '2013-10-1','cc'
union all
select '2013-10-15','dd'
union all
select '2013-10-15','ee'
union all
select '2013-10-20','ff'
 
 
select DENSE_RANK() over(order by s_time) as group_no,
       convert(varchar(10),s_time,120) as s_time,
       na 
from tb
/*
group_no             s_time     na
-------------------- ---------- --------------------
1                    2013-10-01 aa
1                    2013-10-01 bb
1                    2013-10-01 cc
2                    2013-10-15 dd
2                    2013-10-15 ee
3                    2013-10-20 ff
*/

[吃饱饱吖]

谢谢各位，问题解决了，同时也学习到了知识，结贴啦：D