1

输入表查询时差小于或等于1 分钟的所有记录。

输入表

所需输出

输出

>>>>>sqlfiddle <<<<<

4

1 回答 1

3

这是做你想做的吗?

select *
from (select a.*,
             lag(sdt) over (partition by id order by sdt) as prevsdt,
             lead(sdt) over (partition by id order by sdt) as nextsdt
      from table_a a
     ) a
where sdt - prevsdt <= 1/(24*60) or
      nextsdt - sdt <= 1/(24*60);

它在SQL Fiddle上产生所需的输出。

于 2014-07-11T14:59:48.043 回答