mysql获取分组后每组的最大值实例详解

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; 

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

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; 

+----+----------+-------+ | id | name   | score | +----+----------+-------+ | 3 | jason  |   3 | | 6 | linjie  |   3 | | 9 | xiaodeng |   3 | | 11 | hust   |   3 | | 14 | haha   |   2 | | 17 | dengzi  |   5 | | 19 | shazi  |   4 | +----+----------+-------+ 

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;