检查MySQL中日期范围的重叠

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")
;

我们不希望范围之间发生冲突.
假设我们需要从 2010-01-05 到 2010-01-25 插入一个新会话.
我们想知道发生冲突的会话.

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
;

这是结果:

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

有更好的方法吗?

小提琴

答

我曾经用日历应用程序进行过这样的查询.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;

更新这肯定应该有效((ns，ne，es，ee)=(new_start，new_end，existent_start，existent_end)):

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

ns-ne-es-ee:不重叠且不匹配(因为ne< es)
ns-es-ne-ee:重叠且匹配
es-ns-ee-ne:重叠和匹配
es-ee-ns-ne:不重叠且不匹配(因为ns> ee)
es-ns-ne-ee:重叠且匹配
ns-es-ee-ne:重叠和匹配

这里是小提琴

检查MySQL中日期范围的重叠

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