问题描述

此表用于存储会话(事件):

This table is used to store sessions (events):

CREATE TABLE session (
  id int(11) NOT NULL AUTO_INCREMENT
, start_date date
, end_date date
);

INSERT INTO session
  (start_date, end_date)
VALUES
  ("2010-01-01", "2010-01-10")
, ("2010-01-20", "2010-01-30")
, ("2010-02-01", "2010-02-15")
;

我们不想在范围之间发生冲突.
假设我们需要插入一个从 2010-01-05 到 2010-01-25 的新会话.
我们想知道冲突的会话.

We don't want to have conflict between ranges.
Let's say we need to insert a new session from 2010-01-05 to 2010-01-25.
We would like to know the conflicting session(s).

这是我的查询:

SELECT *
FROM session
WHERE "2010-01-05" BETWEEN start_date AND end_date
   OR "2010-01-25" BETWEEN start_date AND end_date
   OR "2010-01-05" >= start_date AND "2010-01-25" <= end_date
;

结果如下:

+----+------------+------------+
| id | start_date | end_date   |
+----+------------+------------+
|  1 | 2010-01-01 | 2010-01-10 |
|  2 | 2010-01-20 | 2010-01-30 |
+----+------------+------------+

有没有更好的方法来获得它?

Is there a better way to get that?

小提琴

推荐答案

我曾经在一个日历应用程序中遇到过这样的问题.我想我使用了这样的东西:

I had such a query with a calendar application I once wrote. I think I used something like this:

... WHERE new_start < existing_end
      AND new_end   > existing_start;

UPDATE 这绝对有效((ns, ne, es, ee) = (new_start, new_end, existing_start, existing_end)):

UPDATE This should definitely work ((ns, ne, es, ee) = (new_start, new_end, existing_start, existing_end)):

1. ns - ne - es - ee:不重叠且不匹配(因为 ne
2. ns - es - ne - ee:重叠和匹配
3. es - ns - ee - ne:重叠和匹配
4. es - ee - ns - ne:不重叠且不匹配(因为 ns > ee)
5. es - ns - ne - ee:重叠和匹配
6. ns - es - ee - ne:重叠和匹配

<小时>

这是一个小提琴

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