根据当天的第一次登录计算员工班次

如何解决根据当天的第一次登录计算员工班次

我想根据他的系统登录时间计算员工班次，我可以根据登录时间计算班次。现在，如果同一天有多次登录，我只想计算给定日期的第一个登录条目的班次。请帮助我如何做到这一点。

**

SQL 查询：

WITH shiftalloence AS
(
       SELECT (timesheet.start_time_server),users.first_name + ' ' + users.last_name AS employee_name,CASE
                     WHEN Dateadd(d,-Datediff(d,dbo.[timesheet].start_time_server),dbo.[timesheet].start_time_server) BETWEEN '05:30:00' AND    '08:00:00' THEN 'Morning'
                     WHEN Dateadd(d,dbo.[timesheet].start_time_server) BETWEEN '08:01:00' AND    '11:00:00' THEN 'General'
                     WHEN Dateadd(d,dbo.[timesheet].start_time_server) BETWEEN '11:05:00' AND    '18:00:00' THEN 'Evening'
                     ELSE '0'
              END AS shift
       FROM   timesheet
       JOIN   users
       ON     timesheet.user_id=users.user_id
       WHERE  Month(start_time_server)=Month(Getdate())
       AND    Datename(weekday,timesheet.start_time_server) IN 'Monday','Tuesday','Wednesday','Thursday','Friday')
AND
start_time_server NOT IN
(
       SELECT timesheet.start_time_server
       FROM   timesheet
       JOIN   leaves
       ON     timesheet.user_id=leaves.user_id
       WHERE  CONVERT(varchar(10),timesheet.start_time_server,111) =CONVERT(varchar(10),leaves.start_time_server,111)
       AND    task_type_name IN ('leave','Half Day') )),tt AS
(
         SELECT   shiftalloence.employee_name,count(
                  CASE
                           WHEN shift = 'Morning' THEN 1
                           ELSE NULL
                  END) AS 'Morning',count(
                  CASE
                           WHEN shift = 'General' THEN 1
                           ELSE NULL
                  END) AS 'General',count(
                  CASE
                           WHEN shift = 'Evening' THEN 1
                           ELSE NULL
                  END) AS 'Evening'
         FROM     shiftalloence
         GROUP BY employee_name )SELECT *
FROM   tt

**

解决方法

您能否根据日期简单地按 ASC 或 DESC 选择和订购？这将允许您仅显示您想要的班次/天/ID，并在他们第一次登录时向您显示，因为您已按照您想要的方式订购。

这是有效的，因为如果您返回 3 行并按日期时间 asc/desc 排序，您也可以简单地选择 top(1)。它只会抓取该轮班/天/人的列表中的最高记录。

我所说的小例子：

SELECT TOP 1 * FROM Employees
order by datetime asc;

类似的东西。