在SQLite中检测日期之间的间隙

Detect gaps between dates in SQLite

提问人:dbz 提问时间:10/17/2023 最后编辑:SelVazidbz 更新时间:10/17/2023 访问量:60

问:

我有一个具有以下结构的SQLite表:

enter image description here

如您所见,该列每 1 分钟有一个新行,但有时该过程会停止,并且存在“时间间隔”,如带下划线的示例所示。我需要通过查询找到这些差距,因为这个表很大。sampleDateSQL

可能的预期结果输出可能是:

2023-03-20 12:56:27
2023-03-29 10:46:46

这可能表示这两个日期之间存在差距。

sql sqlite 间隙和孤岛

评论

1赞 SelVazi 10/17/2023
您能否编辑您的问题并包含一些简单的数据作为文本和预期输出?
0赞 dbz 10/17/2023
@SelVazi,肯定已经添加了!
1赞 SelVazi 10/17/2023
请只用英文写!

答:

1赞 SelVazi 10/17/2023 #1

您可以使用函数检索下一个日期,然后计算差值以检索间隙:LEAD()

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
0赞 Nandalal Seth 10/17/2023 #2

或者您可以使用左反联接来获取没有下一分钟记录的行吗?下面提供的示例 -

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