mysql獲取分組后每組的最大值實(shí)例詳解

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;