問(wèn)題描述
此表用于存儲(chǔ)會(huì)話(huà)(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")
;
我們不想在范圍之間發(fā)生沖突.
假設(shè)我們需要插入一個(gè)從 2010-01-05 到 2010-01-25 的新會(huì)話(huà).
我們想知道沖突的會(huì)話(huà).
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).
這是我的查詢(xún):
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
;
結(jié)果如下:
+----+------------+------------+
| id | start_date | end_date |
+----+------------+------------+
| 1 | 2010-01-01 | 2010-01-10 |
| 2 | 2010-01-20 | 2010-01-30 |
+----+------------+------------+
有沒(méi)有更好的方法來(lái)獲得它?
Is there a better way to get that?
小提琴
推薦答案
我曾經(jīng)在一個(gè)日歷應(yīng)用程序中遇到過(guò)這樣的問(wèn)題.我想我使用了這樣的東西:
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 這絕對(duì)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)):
- ns - ne - es - ee:不重疊且不匹配(因?yàn)?ne
- ns - es - ne - ee:重疊和匹配
- es - ns - ee - ne:重疊和匹配
- es - ee - ns - ne:不重疊且不匹配(因?yàn)?ns > ee)
- es - ns - ne - ee:重疊和匹配
- ns - es - ee - ne:重疊和匹配
<小時(shí)>
這是一個(gè)小提琴
這篇關(guān)于檢查 MySQL 中日期范圍的重疊的文章就介紹到這了，希望我們推薦的答案對(duì)大家有所幫助，也希望大家多多支持html5模板網(wǎng)！