mysql query-one table-find two dates within one day-of the same entity from differ column+diff calculations(mysql查询-一个表-一天内查找两个日期-来自不同列的同一实体+差异计算)

我正试图构造一个mysql查询来计算一个表,其中包含办公室人员的进出口。

我所拥有的:

id  PERSON      IN                  OUT
1   Person A    2019-03-11 08:59:30 NULL
2   Person B    2019-03-11 08:32:00 NULL
3   Person C    2019-03-11 08:04:40 NULL
4   Person D    2019-03-11 07:58:50 NULL
5   Person E    2019-03-11 07:35:20 NULL
6   Person F    2019-03-11 07:35:00 NULL
7   Person A    NULL                2019-03-11 15:00:50
8   Person B    NULL                2019-03-11 14:57:00
8   Person C    NULL                2019-03-11 13:19:50
9   Person D    NULL                2019-03-11 15:14:20
10  Person E    NULL                2019-03-11 15:15:50
11  Person F    NULL                2019-03-11 15:28:10

我想得到的是:

id  PERSON      IN                  OUT                 DIFF IN MINUTES
1   Person A    2019-03-11 08:59:30 2019-03-11 15:00:50 XXX
2   Person B    2019-03-11 08:32:00 2019-03-11 14:57:00 XXX
3   Person C    2019-03-11 08:04:40 2019-03-11 13:19:50 XXX
4   Person D    2019-03-11 07:58:50 2019-03-11 15:14:20 XXX
5   Person E    2019-03-11 07:35:20 2019-03-11 15:15:50 XXX
6   Person F    2019-03-11 07:35:00 2019-03-11 15:28:10 XXX
                                                        TOTAL OF XXXS
                                                        TOTAL OF XXXS - YYY (constant)

这样做的目的是得到一天内在办公室里所花时间的信息。此外,我需要每个人从整个月的会议纪要。每人/每月分组。

我花了一些时间使用这个查询,但效果平平:

SELECT a.PERSON, a.IN, b.OUT, TIMESTAMPDIFF(SECOND,a.IN,b.OUT)-28880 AS WHATSLEFT
FROM presence a
INNER JOIN presence b
ON a.PERSON = b.PERSON
WHERE DATEDIFF(a.IN,b.OUT) = 0 AND b.PERSON ="Person A"
ORDER BY a.IN;

谢谢你的帮助! 亚当