你可以做这样的事情
SELECT empno, date_created, time_in, time_out,
CASE WHEN total_hours - 8 > 0 THEN total_hours - 8 ELSE 0 END overtime
FROM
(
SELECT empno, date_created, time_in, time_out,
TIME_TO_SEC(TIMEDIFF(COALESCE(time_out, '17:00:00'),
COALESCE(time_in, '09:00:00'))) / 3600 total_hours
FROM
(
SELECT empno, date_created,
MIN(CASE WHEN status = 0 THEN time_created END) time_in,
MIN(CASE WHEN status = 1 THEN time_created END) time_out
FROM biometrics
GROUP BY empno, date_created
) a
) b
这是 演示
您需要为提供真正的默认值time_in，并time_out针对当他们NULL。在一个极端的情况下，如果NULLs是由员工有一天到另一天回家而造成的，那么这些默认值可能分别是00:00:00和，23:59:59因为您要计算每个日历日的加班时间。
如果要overtime以时间格式显示
SELECT empno, date_created, time_in, time_out,
SEC_TO_TIME(
CASE WHEN total_sec - 28800 > 0
THEN total_sec - 28800
ELSE 0 END) overtime
FROM
(
SELECT empno, date_created, time_in, time_out,
TIME_TO_SEC(TIMEDIFF(COALESCE(time_out, '17:00:00'),
COALESCE(time_in, '09:00:00'))) total_sec
FROM
(
SELECT empno, date_created,
MIN(CASE WHEN status = 0 THEN time_created END) time_in,
MIN(CASE WHEN status = 1 THEN time_created END) time_out
FROM biometrics
GROUP BY empno, date_created
) a
) b
这是 演示
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