计算 MySQL 中重叠日期范围的最大数量 [英] Count maximum number of overlapping date ranges in MySQL

问题描述

我对以下情况感到头疼.在 MySQL 中，我有一个包含 40000 多个条目的表，如下所示:

I've having a big headache on the following situation. In MySQL I have a table with more than 40000 entries that look like that:

create table if not exists sessions
(
    startt datetime null,
    endt datetime null,
    id int auto_increment
        primary key
);

INSERT INTO sessions (startt, endt, id) VALUES 
('2020-02-06 10:33:55', '2020-02-06 10:34:41', 20356),
('2020-02-06 10:33:14', '2020-02-06 10:33:57', 20355),
('2020-02-06 10:32:55', '2020-02-06 10:33:32', 20354),
('2020-02-06 10:33:03', '2020-02-06 10:33:12', 20353),
('2020-02-06 10:31:38', '2020-02-06 10:32:41', 20352),
('2020-02-06 09:48:44', '2020-02-06 09:50:37', 20351);

SELECT * FROM sessions;
+---------------------+---------------------+-------+
| startt              | endt                | id    |
+---------------------+---------------------+-------+
| 2020-02-06 10:33:55 | 2020-02-06 10:34:41 | 20356 |
| 2020-02-06 10:33:14 | 2020-02-06 10:33:57 | 20355 |
| 2020-02-06 10:32:55 | 2020-02-06 10:33:32 | 20354 |
| 2020-02-06 10:33:03 | 2020-02-06 10:33:12 | 20353 |
| 2020-02-06 10:31:38 | 2020-02-06 10:32:41 | 20352 |
| 2020-02-06 09:48:44 | 2020-02-06 09:50:37 | 20351 |
+---------------------+---------------------+-------+
6 rows in set (0.00 sec)

小提琴 https://www.db-fiddle.com/f/49bNZ7863gv6RThoPpuiid/0

日期和时间范围是会话.我想知道的是:一次存在的最大会话数是多少?

The date and time ranges are sessions. What I want to find out is: what is the maximum number of sessions that existed at one time?

我发现了很多东西，比如如何找出一个日期是否在其他日期的范围内等等.这并没有真正帮助，因为我想知道在最大峰值时有多少用户.

I found a lot of things like how to find out if a date is in the range of other dates etc. which didn't really help as I want to find out how many users there were at the maximum peak.

推荐答案

这里有一个使用窗口函数的选项(在 MySQL 8.0 中可用):

Here is one option using window functions (available in MySQL 8.0):

select dt, sum(nb) over(order by dt) sum_nb
from (
    select starttt dt, 1 nb from mytable 
    union all select endt, -1 from mytable 
) t
order by sum_nb desc
limit 1

这个想法是对数据集进行反透视；并发会话数在每个会话开始时增加 1，并在结束时减少 1.

The idea is to unpivot the dataset; the count of concurrent sessions increases by 1 at the beginning of each session, and decreses by 1 at its end.

然后，您可以使用窗口总和计算每个时间点的并发会话数.

You can then compute the number of concurrent sessions at each point in time with a window sum.

最后一步是按会话数排序并仅保留第一行.

The last step is ordering by session count and keeping the first row only.

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