SQL 在按时间分组的时间范围内有效的行数

Question

SQL 服务器

我有一个带有 2 个时间戳的 table，time_start 和 time_end。例如

ID    time_start           time_end
----  -------------------  -------------------
1     2019-01-01 08:30:00  2019-01-01 09:40:00
2     2019-01-01 09:10:24  2019-01-01 15:14:19
3     2019-01-01 09:21:15  2019-01-01 09:21:19
4     2019-01-01 10:39:45  2019-01-01 10:58:12
5     2019-01-01 11:39:45  2019-01-01 11:40:10

我想对它们进行分组，这样我就可以按可变时间间隔对行数进行分组。例如

time_interval         row_count
-------------------   ---------
2019-01-01 07:00:00   0
2019-01-01 08:00:00   1
2019-01-01 09:00:00   3
2019-01-01 10:00:00   2
2019-01-01 11:00:00   1
2019-01-01 12:00:00   0

我的间隔可以是 1 小时、1 分钟、30 分钟、1 天等...

将其视为 log-in/log-out 情况，我想了解在任何给定的分钟、小时、天等情况下，用户的登录情况...

Answer 1

试试这个，

     DECLARE @start_date datetime='2019-01-01',
                @end_date datetime='2019-01-02',
                @i_minutes int=60

        DECLARE @t TABLE
        (
            id int identity(1,1),time_start datetime,time_end datetime
        )
        INSERT INTO @t(time_start,time_end)VALUES
        ('2019-01-01 08:30:00','2019-01-01 09:40:00'),
        ('2019-01-01 09:10:24','2019-01-01 15:14:19'),
        ('2019-01-01 09:21:15','2019-01-01 09:21:19'),
        ('2019-01-01 10:39:45','2019-01-01 10:58:12'),
        ('2019-01-01 11:39:45','2019-01-01 11:40:10')

        --SELECT @start_date=min(time_start),@end_date=max(time_end)
        --FROM @t

        ;WITH CTE_time_Interval AS
        (
            SELECT @start_date AS time_int,@i_minutes AS i_minutes
            UNION ALL
            SELECT dateadd(minute,@i_minutes,time_int),i_minutes+ @i_minutes
            FROM CTE_time_Interval
            WHERE time_int<=@end_date
        )
        ,CTE1 AS
        (
            SELECT ROW_NUMBER()OVER(ORDER BY time_int)AS r_no,time_int 
            FROM CTE_time_Interval
        )
        ,CTE2 AS
        (
            SELECT a.time_int AS Int_start_time,b.time_int AS Int_end_time 
            FROM CTE1 a
            INNER JOIN CTE1 b ON a.r_no+1=b.r_no
        )

SELECT a.Int_start_time,a.Int_end_time,sum(iif(b.time_start is not null,1,0)) AS cnt
FROM CTE2 a
LEFT JOIN @t b ON
(
    b.time_start BETWEEN a.Int_start_time AND a.Int_end_time 
    OR 
    b.time_end BETWEEN a.Int_start_time AND a.Int_end_time
    OR
    a.Int_start_time BETWEEN b.time_start AND b.time_end
    OR
    a.Int_end_time BETWEEN b.time_start AND b.time_end 
)
GROUP BY a.Int_start_time,a.Int_end_time

Answer 2

嗨，这是我的解决方法。
我用你的数据创建了一个 table "test"。

首先我得到最小和最大间隔，然后我用 CTE 得到这些值之间的所有间隔。最后，通过这个 CTE 和左连接 time_start 和 time_end 之间的间隔，我得到了答案。

这是 1 小时的间隔

DECLARE @minDate AS DATETIME;
DECLARE @maxDate AS DATETIME;

SET @minDate = (select case 
when (select min(time_start) from test) < (select min(time_end) from test) 
then (select min(time_start) from test)
else (select min(time_end) from test) end )
SET @minDate = FORMAT(@minDate, 'dd-MM.yyyy HH:00:00')

SET @maxDate = (select case 
when (select max(time_start) from test) > (select max(time_end) from test) 
then (select max(time_start) from test)
else (select max(time_end) from test) end )
SET @maxDate = FORMAT(@maxDate, 'dd-MM.yyyy HH:00:00')

;WITH Dates_CTE
     AS (SELECT @minDate AS Dates
         UNION ALL
         SELECT Dateadd(hh, 1, Dates)
         FROM   Dates_CTE
         WHERE  Dates < @maxDate)

SELECT d.Dates as time_interval, count(*) as row_count
FROM   Dates_CTE d
LEFT JOIN test t on d.Dates 
between (FORMAT(t.time_start, 'dd-MM.yyyy HH:00:00')) 
and (FORMAT(t.time_end, 'dd-MM.yyyy HH:00:00'))  
GROUP BY d.Dates

对于 10 分钟的间隔，您需要进行一些更改。
首先我格式化日期获取分钟 (dd-MM.yyyy HH:mm:00 instead of dd-MM.yyyy HH:00:00) 在左边的连接中，我接近 time_start 和 time_end 到他们的 10 分钟时间（9:30:00 中的 9:32:00）(dateadd(minute, 10 * (datediff(minute, 0, time_start) / 10), 0)):

DECLARE @minDate AS DATETIME;
DECLARE @maxDate AS DATETIME;

SET @minDate = (select case 
when (select min(time_start) from test) < (select min(time_end) from test) 
then (select min(time_start) from test)
else (select min(time_end) from test) end )
SET @minDate = FORMAT(@minDate, 'dd-MM.yyyy HH:mm:00')

SET @maxDate = (select case 
when (select max(time_start) from test) > (select max(time_end) from test) 
then (select max(time_start) from test)
else (select max(time_end) from test) end )
SET @maxDate = FORMAT(@maxDate, 'dd-MM.yyyy HH:mm:00')

;WITH Dates_CTE
     AS (SELECT @minDate AS Dates
         UNION ALL
         SELECT Dateadd(minute, 10, Dates)
         FROM   Dates_CTE
         WHERE  Dates < @maxDate)

SELECT d.Dates as time_interval, count(*) as row_count
FROM   Dates_CTE d
LEFT JOIN test t on d.Dates 
between dateadd(minute, 10 * (datediff(minute, 0, time_start) / 10), 0) 
and dateadd(minute, 10 * (datediff(minute, 0, time_end) / 10), 0)
GROUP BY d.Dates

最后我得到了 1 小时间隔的结果：

+---------------------+-----------+
|    time_interval    | row_count |
+---------------------+-----------+
| 01/01/2019 08:00:00 |         1 |
| 01/01/2019 09:00:00 |         3 |
| 01/01/2019 10:00:00 |         2 |
| 01/01/2019 11:00:00 |         2 |
| 01/01/2019 12:00:00 |         1 |
| 01/01/2019 13:00:00 |         1 |
| 01/01/2019 14:00:00 |         1 |
| 01/01/2019 15:00:00 |         1 |
+---------------------+-----------+

希望对你有用。

Answer 3

您需要指定时间间隔。其余的是 LEFT JOIN/GROUP BY 或相关子查询：

with dates as (
      select convert(datetime, '2019-01-01 07:00:00') as dt
      union all
      select dateadd(hour, 1, dt)
      from dates
      where dt < '2019-01-01 12:00:00'
     )
select dates.dt, count(t.id)
from dates left join
     t
     on dates.dt < t.time_end and
        dates.dt >= dateadd(hour, 1, t.time_start)
group by dates.dt
order by dates.dt;

如果您有很多数据和很多时间段，您可能会发现它的性能很差。如果是这种情况，请提出 new 问题，并提供有关大小和性能的更多信息。

SQL 在按时间分组的时间范围内有效的行数

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