发布于 2017-10-24 20:33:56 | 85 次阅读 | 评论: 0 | 来源: 网友投递

这里有新鲜出炉的Mysql教程,程序狗速度看过来!

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。


这篇文章主要介绍了 mysql获取分组后每组的最大值实例详解的相关资料,需要的朋友可以参考下

 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; 

感谢阅读,希望能帮助到大家,谢谢大家对本站的支持!



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