如何通过SQL语句获取以下结果
有一个表T如下：
dt(datetime)command(varchar)ci(varchar)colcol2
00:00:01.123...0002223333.......
00:00:02.223start0003233666.......
00:00:03.245...0001237332.......
00:00:05.521failed0002334344.......
00:00:06.223...0005528832.......
00:00:08.612start0001178333.......
00:00:09.727...0002466788.......
00:00:11.113complete0005756890.......
00:00:12.829...0006666444.......
00:00:13.326...NULL.......
00:00:14.223start0005528832.......
00:00:15.612start0001178333.......
00:00:16.727...0002466788.......
00:00:17.113complete0005756890.......
00:00:18.829failed0006666444.......
00:00:19.326...0002334422.......
......
说明：在dt列中时间是递增的且唯一，command列中start和failed/complete(互斥)是一对出现的，也就是出现一个start后面必定会出现一个falied或complete，现在要求找出两个start间中间只出现一次complete的记录（从第一个start开始到第二start结束，不包含第二个start），找出的数据就如上表中蓝色部分，如何用SQL实现？

注：如果表后面的一段数据是以start开始complete结尾也算是要找的数据，也就相当于查找前先在表的最后面添加一条虚拟的start记录

[解决办法]

SQL code

改改SELECT c.* FROM #T AS a    INNER JOIN #T AS b ON a.command=b.command AND a.dt<b.dt AND b.dt=(SELECT min(dt) FROM #T WHERE command=a.command AND dt>a.dt)    INNER JOIN #T AS c ON c.dt >= a.dt AND c.dt<b.dtWHERE a.command='start' AND EXISTS(SELECT 1 FROM #T WHERE [dt] BETWEEN a.dt AND b.dt   HAVING count(CASE WHEN [command] IN(N'complete',N'failed') THEN 1 END)=1 AND COUNT(CASE WHEN [command]=N'complete' THEN 1 end)=1)
[解决办法]
应为>10
SQL code
;with c1 as(select *,rn=ROW_NUMBER()over(order by dt) from tb),c2 as(select *,dt as flg1,0 as flg2 from c1 where command='start'union allselect b.*,a.flg1,(case when b.command='complete' then a.flg2+10 when b.command='failed' then 20 else a.flg2 end)from c2 a inner join c1 b on b.rn=a.rn+1 and b.command<>'start')select dt,command,ci,col,col2 from c2 a where exists(select 1 from c2 where flg1=a.flg1 and flg2=10)and not exists(select 1 from c2 where flg1=a.flg1 and flg2>10)