mysql获取分组后每组的最大值

1. 测试数据库表如下：

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：

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

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:

+----+----------+-------+
| 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

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;