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
有可能这样做吗?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?
答案 0 :(得分:12)
您可以使用PIVOT函数来获取结果,我只是将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;
您还可以使用带有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;
答案 1 :(得分:6)
您无需PIVOT查询即可获取所需信息。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)
而不是TOP 1
。
以下是SQL Fiddle在SQL Server中演示上述查询。