mysql获取分组后每组的最大值实例详解

2023年 4月 28日 37.3k 0

mysql获取分组后每组的最大值实例详解 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

 mysql获取分组后每组的最大值实例详解

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;

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