mysql獲取分組后每組的最大值實例詳解
1. 測試數(shù)據(jù)庫表如下：
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. 插入如下數(shù)據(jù)：
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. 下面是重點，目的是要按照name分組，然后分組后，獲取每組中score分數(shù)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;
當然，上面的最后的order by a.name可以去掉
4. 測試結(jié)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. 網(wǎng)上很多方法都是錯誤的，比如如下一些，親測是不行的
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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