发布于 2017-10-24 12:33:56 | 229 次阅读 | 评论: 0 | 来源: 网友投递
Mysql关系型数据库管理系统
MySQL是一个开放源码的小型关联式数据库管理系统，开发者为瑞典MySQL AB公司。MySQL被广泛地应用在Internet上的中小型网站中。由于其体积小、速度快、总体拥有成本低，尤其是开放源码这一特点，许多中小型网站为了降低网站总体拥有成本而选择了MySQL作为网站数据库。
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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