从下表中查询每个uid最近一条记录:

查询结果group by uid，同一个uid选择id最大那条记录

mysql> select * from test_group;
+----+-----+--------+---------+
| id | uid | sn     | created |
+----+-----+--------+---------+
|  1 |   1 | NUM-01 |  123450 |
|  2 |   2 | NUM-02 |  123451 |
|  3 |   1 | NUM-03 |  123452 |
|  4 |   2 | NUM-04 |  123453 |
|  5 |   1 | NUM-05 |  123454 |
|  6 |   1 | NUM-06 |  123455 |
|  7 |   1 | NUM-07 |  123456 |
|  8 |   3 | NUM-08 |  123457 |
|  9 |   3 | NUM-09 |  123458 |
+----+-----+--------+---------+
9 rows in set (0.05 sec)

• SQL：

select s.*   
from (SELECT max(id) as id FROM `test_group` group by `uid`) t   
left join `test_group` as s on t.id=s.id 

• 查询结果：

+------+------+--------+---------+
| id   | uid  | sn     | created |
+------+------+--------+---------+
|    7 |    1 | NUM-07 |  123456 |
|    4 |    2 | NUM-04 |  123453 |
|    9 |    3 | NUM-09 |  123458 |
+------+------+--------+---------+
3 rows in set (0.00 sec)