SQL Server：pivot功能,需要透视表

我在SQL Server中有以下格式的数据.

-ID          ID2  status  time
-1384904453  417  stop    2013-11-19 23:40:43.000
-1384900211  417  start   2013-11-19 22:30:06.000
-1384822614  417  stop    2013-11-19 00:56:36.000
-1384813810  417  start   2013-11-18 22:30:06.000
-1384561199  417  stop    2013-11-16 00:19:45.000
-1384554623  417  start   2013-11-15 22:30:06.000
-1384475231  417  stop    2013-11-15 00:26:58.000
-1384468224  417  start   2013-11-14 22:30:06.000
-1384388181  417  stop    2013-11-14 00:16:20.000
-1384381807  417  start   2013-11-13 22:30:06.000
-1384300222  417  stop    2013-11-12 23:50:11.000
-1384295414  417  start   2013-11-12 22:30:06.000
-1384218209  417  stop    2013-11-12 01:03:17.000
-1384209015  417  start   2013-11-11 22:30:06.000

我需要的是能够以下列格式显示数据.

-ID2  start                      stop
-417  2013-11-19 22:30:06.000    2013-11-19 23:40:43.000
-417  2013-11-18 22:30:06.000    2013-11-19 00:56:36.000

是否有可能做到这一点？我在SQL Server中尝试过pivot,但它只返回一条记录.有人可以帮忙吗？

解决方法

select id2,start,stop
from
(
  select id2,status,time,row_number() over(partition by status
                      order by time) seq
  from yourtable
) d
pivot
(
  max(time)
  for status in (start,stop)
) piv
order by start desc;

见SQL Fiddle with Demo.

您还可以使用带有CASE表达式的聚合函数来获取最终结果：

select
  id2,max(case when status = 'start' then time end) start,max(case when status = 'start' then time end) stop
from 
(
  select id2,row_number() over(partition by status
                      order by time) seq
  from yourtable
) d
group by id2,seq;

见SQL Fiddle with Demo

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