从输入表查询时差小于或等于1 分钟的所有记录。
所需输出
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。