SQL Server:透视功能，需要透视一个表

Question

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

我需要的是能够以以下格式显示数据。

-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

有可能做到这一点吗？我尝试在SQL Server中使用pivot，但它只返回一条记录。有人能帮忙吗？

Answer 1

您可以使用PIVOT函数来获得结果，我只需要将row_number()窗口函数应用于数据，这样您就可以为每个ID2:返回多行

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;

请参阅SQL Fiddle with Demo。

您也可以使用带有CASE表达式的聚合函数来获得最终结果：

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

请参阅带有演示的SQL Fiddle

Answer 2

您不需要PIVOT查询就可以获得所需的信息。您可以执行以下操作：

SELECT mt1.ID2, mt1.time AS start,
(
   SELECT TOP 1 mt2.time
   FROM MyTable AS mt2
   WHERE mt2.status = 'stop' 
     AND mt2.time >= mt1.time
   ORDER BY mt2.time 
) AS stop
FROM MyTable AS mt1
WHERE mt1.status = 'start' 

如果您在SQL Server而不是MS Access中执行上述查询，则需要使用TOP(1)而不仅仅是TOP 1。

以下是SQL Fiddle在SQL Server中演示上述查询。