SQL在特定条件下计算时间差

Question

我在 SQL 服务器中有一个表，其中的数据如下图所示。所以我需要计算列值大于或等于 3 的日期与下一列值 0 之间的时间差。所以在下面的示例中，“9:51”，值为 3，在“9: 54"。所以这里的总时差是3分钟。再次下一个时段是从“9:57”到“10:02”，所以这里的时间差是5分钟。下一个插槽来自“10:03”，但没有对应的行具有 0 值。所以这里的时差是 4 分钟（从“10:03”到“10:06”）。所以总时差是（3+5+4），12分钟。任何形式的帮助将不胜感激。提前致谢。

   DateTime    Value
3/10/2020 9:50  1
3/10/2020 9:51  3
3/10/2020 9:52  1
3/10/2020 9:53  2
3/10/2020 9:54  0
3/10/2020 9:55  0
3/10/2020 9:56  1
3/10/2020 9:57  3
3/10/2020 9:58  2
3/10/2020 9:59  3
3/10/2020 10:00 2
3/10/2020 10:01 2
3/10/2020 10:02 0
3/10/2020 10:03 3
3/10/2020 10:04 1
3/10/2020 10:05 1
3/10/2020 10:06 1

编辑 2：我也有一个雪花数据库，并使用我在上表中生成的以下查询。

   with aggregatedData as 
  (
       select count ( EVENT_ID) as total , time_slice(to_timestamp_ntz(EVENT_CREATED_AT), 1, 'MINUTE') as minutePeriod
            from events  
            group by minutePeriod

            union

            SELECT
            0 as total,
            DATEADD(MINUTE, SEQ4(), time_slice(to_timestamp_ntz(select min(EVENT_CREATED_AT) from events), 1, 'MINUTE')) AS minutePeriod
            FROM TABLE(GENERATOR(ROWCOUNT=>20000))
            WHERE minutePeriod <= time_slice(to_timestamp_ntz(select max(EVENT_CREATED_AT) from events), 1, 'MINUTE')
 )
 select minutePeriod
  ,sum(total) over (partition by null order by minutePeriod rows between 9 preceding and current row) totalValue
  from aggregatedData
  order by minutePeriod;

我期待低于输出

Answer 1

你可以试试这个。

SELECT T1.*, 
    CASE 
        WHEN T1.Value = 3 AND (X3.Value = 0 OR X3.Value IS NULL) 
        THEN DATEDIFF(MINUTE,T1.DateTime , ISNULL(X0.[DateTime],  (SELECT DATEADD(MINUTE,1, MAX([DateTime])) FROM SampleData) )) END
FROM SampleData T1 
    OUTER APPLY ( SELECT TOP 1 * FROM SampleData T2 WHERE T2.[DateTime] > T1.[DateTime] AND T2.Value = 0 ORDER BY T2.DateTime ASC) X0
    OUTER APPLY ( SELECT TOP 1 * FROM SampleData T2 WHERE T2.[DateTime] < T1.[DateTime] AND T2.Value IN (0, 3) ORDER BY T2.[DateTime] DESC) X3
ORDER BY T1.[DateTime]

结果：

DateTime                Value       
----------------------- ----------- -----------
2020-03-10 09:50:00.000 1           NULL
2020-03-10 09:51:00.000 3           3
2020-03-10 09:52:00.000 1           NULL
2020-03-10 09:53:00.000 2           NULL
2020-03-10 09:54:00.000 0           NULL
2020-03-10 09:55:00.000 0           NULL
2020-03-10 09:56:00.000 1           NULL
2020-03-10 09:57:00.000 3           5
2020-03-10 09:58:00.000 2           NULL
2020-03-10 09:59:00.000 3           NULL
2020-03-10 10:00:00.000 2           NULL
2020-03-10 10:01:00.000 2           NULL
2020-03-10 10:02:00.000 0           NULL
2020-03-10 10:03:00.000 3           4
2020-03-10 10:04:00.000 1           NULL
2020-03-10 10:05:00.000 1           NULL
2020-03-10 10:06:00.000 1           NULL

Answer 2

所以我需要计算列值大于或等于 3 的日期与下一列值为 0 的日期之间的时间差。

如果我理解正确：

select t.*,
       (case when value >= 3
             then datediff(minute,
                           datetime,
                           min(case when value = 0 then datetime end) over (order by datetime desc)
                          )
        end) as diff_minutes
from t;

Here 是一个 dbfiddle。