MYSQL - 如何在同一个表上的两个不同行上减去两列？
[英]MYSQL - How may i subtract two columns on two different rows, on the same table?
问题描述
I have a select taht brings the result like this:
我有一个选择 taht 带来这样的结果：
+-----------+------------+--------------+
| parking_id| start_time | end_time |
+-----------+------------+--------------+
| 38 | 09:15:00 | 10:32:00 |
| 57 | 11:45:00 | 13:21:00 |
| 33 | 14:40:00 | 16:35:00 |
| 15 | 17:13:00 | 19:15:00 |
| 68 | 20:54:00 | NULL |
+-----------+------------+--------------+
As you can see the IDs dont follow a linear order, but wat i really need is a select that brings me the time between the new start_time and end_time for the last inserted , that follows this non linear order, so i need a select that brings me this:正如您所看到的，ID 不遵循线性顺序，但是我真正需要的是一个选择，它为我带来最后一次插入的新start_time和end_time之间的时间，遵循这个非线性顺序，所以我需要一个选择我这个：
+-----------+------------+--------------+----------------+
| parking_id| start_time | end_time | time_btw_parks |
+-----------+------------+--------------+----------------+
| 38 | 09:15:00 | 10:32:00 | NULL |
| 57 | 11:45:00 | 13:21:00 | 01:13:00 |
| 33 | 14:40:00 | 16:35:00 | 01:19:00 |
| 15 | 17:13:00 | 19:15:00 | 00:38:00 |
| 68 | 20:54:00 | NULL | 01:39:00 |
+-----------+------------+--------------+----------------+
Doesn't have to necessary be select query.不一定非要选择查询。 Anything that solves it would help.任何解决它的方法都会有所帮助。
1 个解决方案
解决方案1
1 已采纳 2020-02-01 22:48:23
For this sample data you can use timediff() function:对于此示例数据，您可以使用timediff()函数：
select t.parking_id, t.start_time, t.end_time,
timediff(t.start_time, max(tt.end_time)) time_btw_parks
from tablename t left join tablename tt
on t.start_time > tt.end_time
group by t.parking_id, t.start_time, t.end_time
order by t.start_time
See the demo .请参阅演示。
Results:结果：
| parking_id | start_time | end_time | time_btw_parks |
| ---------- | ---------- | -------- | -------------- |
| 38 | 09:15:00 | 10:32:00 | |
| 57 | 11:45:00 | 13:21:00 | 01:13:00 |
| 33 | 14:40:00 | 16:35:00 | 01:19:00 |
| 15 | 17:13:00 | 19:15:00 | 00:38:00 |
| 68 | 20:54:00 | | 01:39:00 |
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