1. 测试数据库表如下：
|
1
2
3
4
5
6
7
|
create table test
(
`id` int not null auto_increment,
`name` varchar(20) not null default '',
`score` int not null default 0,
primary key(`id`)
)engine=InnoDB CHARSET=UTF8;
|
2. 插入如下数据：
|
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
|
mysql> select * from test;
+----+----------+-------+
| id | name | score |
+----+----------+-------+
| 1 | jason | 1 |
| 2 | jason | 2 |
| 3 | jason | 3 |
| 4 | linjie | 1 |
| 5 | linjie | 2 |
| 6 | linjie | 3 |
| 7 | xiaodeng | 1 |
| 8 | xiaodeng | 2 |
| 9 | xiaodeng | 3 |
| 10 | hust | 2 |
| 11 | hust | 3 |
| 12 | hust | 1 |
| 13 | haha | 1 |
| 14 | haha | 2 |
| 15 | dengzi | 3 |
| 16 | dengzi | 4 |
| 17 | dengzi | 5 |
| 18 | shazi | 3 |
| 19 | shazi | 4 |
| 20 | shazi | 2 |
+----+----------+-------+
|
3. 下面是重点，目的是要按照name分组，然后分组后，获取每组中score分数最多的，sql如下
|
1
2
|
select a.* from test a inner join (select name,max(score) score from test group by name)b on a.
name=b.name and a.score=b.score order by a.name;
|
当然，上面的最后的order by a.name可以去掉
4. 测试结果如下:
|
1
2
3
4
5
6
7
8
9
10
11
|
+----+----------+-------+
| id | name | score |
+----+----------+-------+
| 3 | jason | 3 |
| 6 | linjie | 3 |
| 9 | xiaodeng | 3 |
| 11 | hust | 3 |
| 14 | haha | 2 |
| 17 | dengzi | 5 |
| 19 | shazi | 4 |
+----+----------+-------+
|
5. 网上很多方法都是错误的，比如如下一些，亲测是不行的
|
1
2
3
4
5
6
7
8
9
10
11
|
select * from (select * from test order by score desc) t group by name order by score desc limit 4;
select score,max(score) from test group by name;
select * from test where score in (select max(score) from test group by name);
select * from test where score in (select substring_index(group_concat(score order by score desc separator ','),',',1) from test group by name);
select * from (select name,score,ROW_NUMBER() over(group by name order by score desc) as rowNum from test) rank where rank.rowNum <=1 order by rank.score desc;
select * from( select StoresNo,[CustomerCaseNo],[PaymentsTime], ROW_NUMBER() over(partition by CustomerCaseNo order by [PaymentsTime] desc) as rowNum
from BAL_paymentsSwiftInfo where StoresNo='zq00000034') ranked where ranked.rowNum <= 1 order by ranked.CustomerCaseNo, ranked.PaymentsTime desc
select * from (select * from test order by score desc) as a group by a.name;
|
感谢阅读，希望能帮助到大家，谢谢大家对本站的支持！
原文链接：http://blog.csdn.net/u011734144/article/details/51982134