1. 测试数据库表如下：
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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：
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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
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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:
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+----+----------+-------+ | 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
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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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原文链接：http://blog.csdn.net/u011734144/article/details/51982134
