我使用最大-n-group-group 标记阅读了很多问题，但我找不到比此{1}}技巧更好的解决方案：< / p>
UNION ALL我想为五个不同的小组选择十个最长的工作（ GRP01％， GRP04％， GRP12％， GRP15 ％和 GRP20％）。
我使用的是SQLite
select * from (select GroupName, JobName, Start, End, Status, (strftime('%s', End) - strftime('%s', Start)) as Length from ReportJobs where PlanDate = '2014-02-13' and GroupName like 'GRP01%' ORDER BY Length DESC LIMIT 10)
UNION ALL
select * from (select GroupName, JobName, Start, End, Status, (strftime('%s', End) - strftime('%s', Start)) as Length from ReportJobs where PlanDate = '2014-02-13' and GroupName like 'GRP04%' ORDER BY Length DESC LIMIT 10)
UNION ALL
select * from (select GroupName, JobName, Start, End, Status, (strftime('%s', End) - strftime('%s', Start)) as Length from ReportJobs where PlanDate = '2014-02-13' and GroupName like 'GRP12%' ORDER BY Length DESC LIMIT 10)
UNION ALL
select * from (select GroupName, JobName, Start, End, Status, (strftime('%s', End) - strftime('%s', Start)) as Length from ReportJobs where PlanDate = '2014-02-13' and GroupName like 'GRP15%' ORDER BY Length DESC LIMIT 10)
UNION ALL
select * from (select GroupName, JobName, Start, End, Status, (strftime('%s', End) - strftime('%s', Start)) as Length from ReportJobs where PlanDate = '2014-02-13' and GroupName like 'GRP20%' ORDER BY Length DESC LIMIT 10);
和Start列中的元素是ISO 8601格式
我使用End来计算每个作业的长度，然后使用(strftime('%s', End) - strftime('%s', Start)) as Length来计算每个作业。
您是否知道执行此查询的最简单/更好的方法（使用SQLite）？