Question

SELECT empno , date_created , 
min(CASE WHEN status = 0 THEN time_created END) time_in,
max(CASE WHEN status = 1 THEN time_created END) time_out
FROM biometrics
WHERE empno = 3
GROUP BY empno , date_created

在此语法中，它仅确定员工的第一个和最后一个但是我想问一下如何计算他/她在员工的第一个IN和最后一个之间的时差或总时间。

例如员工的第一个上午9:00:00，最后一个下午是16:00:00。如何计算该员工的总工时？

Answer 1

请尝试以下方法：

select 
  empno, 
  date_created, 
  time_in, 
  time_out, 
  time_format(timediff(time_out, time_in), '%H:%i') as total_time 
from 
(
  SELECT empno, date_created, 
    min(CASE WHEN status = 0 THEN time_created END) time_in,
    max(CASE WHEN status = 1 THEN time_created END) time_out
  FROM biometrics
  WHERE empno = 3 
  GROUP BY empno, date_created
) t1;

Answer 2

通过这个你可以找到firstin和lastout之间的小时差异


SELECT DATEDIFF(hour, firstin(datetime), lastout(datetime)) FROM table_name

计算MySQL中的时差

2 个答案: