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  • Mysql查找所有项目开始时间比之前项目结束时间小的项目ID

    这是之前遇到过的一道sql面试题,供参考学习:

    查找所有项目开始时间比之前项目结束时间小的项目ID

    mysql> select * from t2;
    +----+---------------------+---------------------+
    | id | start               | end                 |
    +----+---------------------+---------------------+
    |  1 | 2015-01-01 00:00:00 | 2015-01-29 00:00:00 |
    |  2 | 2015-01-01 00:00:00 | 2015-01-22 00:00:00 |
    |  3 | 2015-01-03 00:00:00 | 2015-01-20 00:00:00 |
    |  4 | 2015-02-03 00:00:00 | 2015-03-20 00:00:00 |
    |  5 | 2015-01-04 00:00:00 | 2015-01-15 00:00:00 |
    |  6 | 2015-01-18 00:00:00 | 2015-02-15 00:00:00 |
    +----+---------------------+---------------------+

    数据如上所示,由人工观察可知id为2,3,5,6的数据列将需要被找出。

    sql语句如下:

    mysql> select distinct(a.id) from t2 a,t2 b where a.start<b.end and a.id>b.id;
    +----+
    | id |
    +----+
    |  2 |
    |  3 |
    |  5 |
    |  6 |
    +----+
    4 rows in set (0.00 sec)
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  • 原文地址:https://www.cnblogs.com/zejin2008/p/4714849.html
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