Question

我正在努力了解员工迟到的频率

到目前为止我已经使用了这个

SELECT  EMPID 
       ,EMPNAME
      ,COUNT(TimeIn) AS LATECOUNT  FROM        
 (SELECT  EMPLOYEEID AS EMPID, EmployeeName AS EMPNAME, AbsenceDate AS ABSDAT, MIN(TimeIn) AS TimeIn
   FROM  dbo.viewAttendance
         GROUP BY AbsenceDate, EMPLOYEEID, EmployeeName
         HAVING      (MIN(TimeIn) > '08:00:00')
  ) AS a

使用该代码，我只能检索只有员工迟到至少一行（仅返回30行）的行。我真正想要的是300行（所有员工）的完整回报，显示他们迟到的人数。我的想法不够明确，无法找到正确的语法。我知道问题是在HAVING（MIN（timeIn）＆gt; '08：00：00'有人可以帮助我吗？在此先感谢... :)

Answer 1

SELECT
  E.EMPNAME
  ISNULL(SUM(CASE WHEN A.TIMEIN > '08:00:00' THEN 1 ELSE 0 END), 0) AS LATE_ATTENDANCE
FROM
 dbo.MASTEREMPLOYEE E
 LEFT JOIN dbo.viewATTENDANCE A ON A.EMPLOYEEID = E.EMPLOYEEID

Answer 2

为什么不使用这样的东西：

select e.empname, (select count(1) from dbo.viewAttendance a where a.EmployeeId = e.employeeId and a.timein > '08:00:00') as latecount
from employees e

迟到的查询语法

2 个答案: