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)