Question

我有呼叫中心座席的登录和注销表，这些座席的工作班次是从22到06，所以我需要计算的是loginTime的最小值和logOutTime的最大值，但是这变得很复杂，因为在某些情况下，在该时间段内业务代表断开连接，然后在另一个轮班开始的同一天在数据库中创建另一个登录事件，依此类推。

我尝试使用ROW_NUMBER函数以及将MIN和MAX与ROWS PRECEDING和FOLLOWING一起使用来管理结果，但是它不起作用，我的代码：

SELECT  agentId
       ,position
       ,loginDate
       ,loginTime
       ,logoutDate
       ,logoutTime
       ,MIN((CAST(loginDate AS DATETIME) + CAST(loginTime AS DATETIME)))
             OVER(PARTITION BY agentId, position ORDER BY agentId, loginDate, loginTime) minLoginTime
       ,MAX((CAST(logoutDate AS DATETIME) + CAST(logoutTime AS DATETIME)))
       OVER(PARTITION BY agentId, position ORDER BY agentId, loginDate, loginTime ROWS BETWEEN 0 PRECEDING AND 1 FOLLOWING) maxLogoutTime
FROM @tbl
ORDER BY loginDate
       ,loginTime
       ,logoutDate
       ,logoutTime

这是表格示例：

DECLARE @tbl TABLE(agentId VARCHAR(10), position VARCHAR(10), loginDate DATE, loginTime TIME(0), logoutDate DATE, logoutTime TIME(0))

INSERT INTO @tbl SELECT '311338', '230025', '2019-06-03', '21:59:00', '2019-06-04', '06:00:00'
INSERT INTO @tbl SELECT '311338', '230025', '2019-06-04', '21:59:00', '2019-06-04', '23:30:00'
INSERT INTO @tbl SELECT '311338', '230025', '2019-06-04', '23:31:00', '2019-06-05', '06:01:00'
INSERT INTO @tbl SELECT '311338', '230038', '2019-06-05', '21:59:00', '2019-06-06', '02:13:00'
INSERT INTO @tbl SELECT '311338', '230038', '2019-06-06', '02:14:00', '2019-06-06', '06:00:00'
INSERT INTO @tbl SELECT '311338', '230037', '2019-06-06', '22:00:00', '2019-06-07', '06:00:00'
INSERT INTO @tbl SELECT '311338', '230038', '2019-06-07', '21:59:00', '2019-06-08', '00:53:00'
INSERT INTO @tbl SELECT '311338', '230038', '2019-06-08', '00:53:00', '2019-06-08', '06:00:00'
INSERT INTO @tbl SELECT '311338', '230038', '2019-06-09', '22:00:00', '2019-06-10', '06:09:00'

SELECT agentId
    ,position
    ,(CAST(loginDate AS DATETIME) + CAST(loginTime AS DATETIME)) loginTime
    ,(CAST(logoutDate AS DATETIME) + CAST(logoutTime AS DATETIME)) logoutTime
  FROM @tbl

agentId 311338的预期结果将是：

2019年5月6日：minLogin = 2019-06-05 21：59：00.000 maxLogout = 2019-06-06 06：00：00.000

2019年6月6日：minLogin = 2019-06-06 22：00：00.000 maxLogout = 2019-06-07 06：00：00.000

2019年6月7日：minLogin = 2019-06-07 21：59：00.000 maxLogout = 2019-06-08 06：00：00.000

一天。...

如您所见，没有时间可以浪费，如果同一天有多个登录事件，我必须检查它是否属于上一天班次或属于第二天班次。

希望你们能帮助我。

Answer 1

只要人们在08:00之后不停止轮班，这对您就会有用（为简单起见，我稍微更改了您的餐桌）

DECLARE @tbl TABLE(agentId VARCHAR(10), loginMoment DATETIME, logoutMoment DATETIME)

INSERT INTO @tbl SELECT '311338', '2019-06-03 21:59:00', '2019-06-04 06:00:00'
INSERT INTO @tbl SELECT '311338', '2019-06-04 21:59:00', '2019-06-04 23:30:00'
INSERT INTO @tbl SELECT '311338', '2019-06-04 23:31:00', '2019-06-05 06:01:00'
INSERT INTO @tbl SELECT '311338', '2019-06-05 21:59:00', '2019-06-06 02:13:00'
INSERT INTO @tbl SELECT '311338', '2019-06-06 02:14:00', '2019-06-06 06:00:00'
INSERT INTO @tbl SELECT '311338', '2019-06-06 22:00:00', '2019-06-07 06:00:00'
INSERT INTO @tbl SELECT '311338', '2019-06-07 21:59:00', '2019-06-08 00:53:00'
INSERT INTO @tbl SELECT '311338', '2019-06-08 00:53:00', '2019-06-08 06:00:00'
INSERT INTO @tbl SELECT '311338', '2019-06-09 22:00:00', '2019-06-10 06:09:00'


SELECT agentId
    ,DATEADD(HOUR, 8, MIN(DATEADD(HOUR, -8, loginMoment)))
    ,DATEADD(HOUR, 8, MAX(DATEADD(HOUR, -8, logoutMoment)))
FROM @tbl
GROUP BY agentId, CAST(DATEADD(HOUR, -8, loginMoment) AS DATE)

结果如您所愿：

agentId login logout
311338  03/06/2019 21:59:00 04/06/2019 06:00:00
311338  04/06/2019 21:59:00 05/06/2019 06:01:00
311338  05/06/2019 21:59:00 06/06/2019 06:00:00
311338  06/06/2019 22:00:00 07/06/2019 06:00:00
311338  07/06/2019 21:59:00 08/06/2019 06:00:00
311338  09/06/2019 22:00:00 10/06/2019 06:09:00

Answer 2

据我所知，您感兴趣的“一天”比正常日期晚6小时-因此，请在“日期/时间”之后6小时尝试“分组”。

为了进一步说明，我将登录时间减去6小时，然后将日期部分按（名义日期）分组

    SELECT  agentId, 
        CAST(DATEADD(hour,-6,CAST(loginDate AS DATETIME) + CAST(loginTime AS DATETIME)) AS DATE) NotionalDate
        ,min((CAST(loginDate AS DATETIME) + CAST(loginTime AS DATETIME))) minloginTime
        ,max((CAST(logoutDate AS DATETIME) + CAST(logoutTime AS DATETIME))) maxlogoutTime
    FROM @tbl 
        group by agentId, 
            CAST(DATEADD(hour,-6,CAST(loginDate AS DATETIME) + CAST(loginTime AS DATETIME)) AS DATE)

使用登录事件计算SQL Server中登录和注销时间的最小和最大

2 个答案: