每 2 行以分钟为单位的 DATEDIFF 总和

Question

我正在尝试针对第 3 方员工时间跟踪数据库运行查询。据我所知，他们并没有全天记录员工时间的总和。我所拥有的是一系列包含用户 ID 和时间戳的行。如果我按时间戳对行进行排序，我可以有效地获得他们的出拳历史。如果我假设打孔 1 进入，打孔 2 打出，打孔 3 进入等等，是否有一种有效的方法可以在几分钟内从每隔一行找到 DATEDIFF，然后将它们相加以获得当天的总时间给那个员工？

    badge_no punch_timestamp
    11209   1/31/14 7:58 AM
    11209   1/31/14 9:57 AM
    11209   1/31/14 10:00 AM
    11209   1/31/14 10:07 AM

在我发帖后不到 2 分钟，我看到了这篇文章： SQL Server find datediff between different rows, sum

我会先试一试。

Answer 1

这是一种相对简单、幼稚的做法。假设，就像你说的那样，每个“奇数”行是一个印记，每个“偶数”行是一个印记，你可以分别获取奇数行和偶数行并计算每个工作块。请注意，我使用的 DateDiff 以分钟为单位 (mi) 但您可以将其更改为小时/秒/任意值：http://technet.microsoft.com/en-us/library/ms189794.aspx

;WITH StartTime AS
(
SELECT
    badge_no,
    punch_timestamp,
    myrow
FROM
(
    SELECT
        badge_no,
        punch_timestamp,
        ROW_NUMBER() OVER (Partition BY badge_no ORDER BY punch_timestamp ASC) as myrow
    FROM #Time
) [t1]
WHERE myrow % 2 = 1 --odd rows
)
,EndTime AS
(
SELECT
    badge_no,
    punch_timestamp,
    myrow - 1 as 'myrow' --Subtract 1 to match up with the odd rows
FROM
(
    SELECT
        badge_no,
        punch_timestamp,
        ROW_NUMBER() OVER (Partition BY badge_no ORDER BY punch_timestamp ASC) as myrow
    FROM #Time
) [t1]
WHERE myrow % 2 = 0 --even rows
)

SELECT
    badge_no,
    SUM(diff) as 'MinutesWorked'
FROM 
(
SELECT
    EndTime.badge_no,
    DATEDIFF(mi, 
             (SELECT TOP 1 
                  punch_timestamp 
              FROM StartTime 
              WHERE StartTime.badge_no = EndTime.badge_no 
                  AND StartTime.myrow = EndTime.myrow), 
             EndTime.punch_timestamp) as 'diff'
FROM EndTime
) [t1]
GROUP BY badge_no

这是我使用的测试数据：

CREATE TABLE #Time
(
    badge_no nvarchar(10),
    punch_timestamp datetime
)

INSERT INTO #Time VALUES ('100', '2013-01-02 12:01 PM')
INSERT INTO #Time VALUES ('100', '2013-01-02 1:38 PM')
INSERT INTO #Time VALUES ('100', '2013-01-02 2:29 PM')
INSERT INTO #Time VALUES ('100', '2013-01-03 3:01 PM')
INSERT INTO #Time VALUES ('100', '2013-01-03 4:20 PM')
INSERT INTO #Time VALUES ('100', '2013-01-04 12:01 PM')
INSERT INTO #Time VALUES ('100', '2013-01-04 2:01 PM')
INSERT INTO #Time VALUES ('100', '2013-01-04 3:11 PM')
INSERT INTO #Time VALUES ('100', '2013-01-04 4:21 PM')
INSERT INTO #Time VALUES ('100', '2013-01-05 12:01 PM')
INSERT INTO #Time VALUES ('100', '2013-01-05 1:01 PM')
INSERT INTO #Time VALUES ('200', '2013-01-04 2:11 AM')
INSERT INTO #Time VALUES ('200', '2013-01-04 4:34 PM')
INSERT INTO #Time VALUES ('200', '2013-01-05 1:01 AM')
INSERT INTO #Time VALUES ('200', '2013-01-05 4:29 AM')

Answer 2

使用@DaveZych 示例数据，我已经设法计算出与他相同的结果，使用下面的 SQL 语句：

;WITH DataSource ([StartOrEnd], [badge_no], [punch_timestamp]) AS
(
    SELECT ROW_NUMBER() OVER (PARTITION BY [badge_no] ORDER BY [punch_timestamp]) +
           ROW_NUMBER() OVER (PARTITION BY [badge_no] ORDER BY [punch_timestamp])  % 2
          ,[badge_no]
          ,[punch_timestamp]
    FROM #Time
),
TimesPerBadge_No ([badge_no], [StartOrEnd], [Minutes]) AS
(
    SELECT  [badge_no]
           ,[StartOrEnd] 
           ,DATEDIFF(MINUTE, MIN([punch_timestamp]), MAX([punch_timestamp]))
    FROM DataSource
    GROUP BY [badge_no]
            ,[StartOrEnd] 
)
SELECT [badge_no]
      ,SUM([Minutes])
FROM TimesPerBadge_No
GROUP BY [badge_no]

---

这里可以看到每个CTE的值：

首先，我们需要对每个开始和结束日期进行分组：

 SELECT ROW_NUMBER() OVER (PARTITION BY [badge_no] ORDER BY [punch_timestamp]) +
           ROW_NUMBER() OVER (PARTITION BY [badge_no] ORDER BY [punch_timestamp])  % 2
          ,[badge_no]
          ,[punch_timestamp]
    FROM #Time

现在，我们可以计算每组的分钟差：

SELECT  [badge_no]
        ,[StartOrEnd] 
        ,DATEDIFF(MINUTE, MIN([punch_timestamp]), MAX([punch_timestamp]))
FROM DataSource
GROUP BY [badge_no]
        ,[StartOrEnd] 

最后总结每个badge_no的分钟数：

SELECT [badge_no]
      ,SUM([Minutes])
FROM TimesPerBadge_No
GROUP BY [badge_no]