您可以使用LEAD()函数来检索下一个日期，然后计算差值来检索间隙：

with cte as (
  select *, 
          lead(sampleDate, 1, sampleDate) over (order by sampleDate) as nextDate,
          CAST(strftime('%M', lead(sampleDate, 1, sampleDate) over (order by sampleDate)) as integer) 
          - CAST(strftime('%M', sampleDate) as integer) as diff
  from mytable
)
select sampleDate, nextDate
from cte
where diff > 1

或者你可以使用一个左反连接来获取没有下一分钟记录的行吗？示例如下-

with cte as (
  select 4556 id, datetime('2023-03-20 12:54:27') sampleDate
  union ALL
  select 4557 id, datetime('2023-03-20 12:55:27') sampleDate
  union ALL
  select 4558 id, datetime('2023-03-20 12:56:27') sampleDate
  union ALL
  select 4559 id, datetime('2023-03-29 10:46:46') sampleDate
  union ALL
  select 4560 id, datetime('2023-10-06 10:52:47') sampleDate
  union ALL
  select 4561 id, datetime('2023-10-06 10:53:47') sampleDate
 )
 select a.* from cte a 
 left join cte b ON  b.sampledate = datetime(a.sampleDate,'+1 minutes') -- if wanted the gap to be exact one minutes
            --  ON b.sampledate between datetime(a.sampleDate,'+1 seconds')  and datetime(a.sampleDate,'+1 minutes')  -- if you wanted the gap to be anywhere under a minute
 where b.id is NULL