MySQL分类排名和分组TOP N实例详解

表结构

学生表如下：

create table `t_student` (
  `id` int not null auto_increment,
  `t_id` int default null comment '学科id',
  `score` int default null comment '分数',
  primary key (`id`)
);

数据如下：

题目一：获取每个科目下前五成绩排名（允许并列）

允许并列情况可能存在如4、5名成绩并列情况，会导致取前4名得出5条数据，取前5名也是5条数据。

select
	s1.* 
from
	student s1
	left join student s2 on s1.t_id = s2.t_id 
	and s1.score < s2.score 
group by
	s1.id
having
	count( s2.id ) < 5 
order by
	s1.t_id,
	s1.score desc

ps：取前4名时

分析：

1.自身左外连接，得到所有的左边值小于右边值的集合。以t_id=1时举例，24有5个成绩大于他的（74、64、54、44、34），是第6名，34只有4个成绩大于他的，是第5名……74没有大于他的，是第一名。

select
	* 
from
	student s1
	left join student s2 on s1.t_id = s2.t_id 
	and s1.score < s2.score

2. 把总结的规律转换成sql表示出来，就是group by 每个student 的 id(s1.id)，having统计这个id下面有多少个比他大的值(s2.id)

select
	s1.* 
from
	student s1
	left join student s2 on s1.t_id = s2.t_id 
	and s1.score < s2.score 
group by
	s1.id
having
	count( s2.id ) < 5

3. 最后根据 t_id 分类，score 倒序排序即可。

题目二：获取每个科目下最后两名学生的成绩平均值

取最后两名成绩

select
	s1.* 
from
	student s1
	left join student s2 on s1.t_id = s2.t_id 
	and s1.score > s2.score 
group by
	s1.id 
having
	count( s1.id )< 2 
order by
	s1.t_id,
	s1.score

并列存在情况下可能导致筛选出的同一t_id 下结果条数大于2条，但题目要求是取最后两名的平均值，多条平均后还是本身，故不必再对其处理，可以满足题目要求。

分组求平均值：

select
	t_id,avg(score)
from
	(
	select
		s1.*
	from
		student s1
		left join student s2 on s1.t_id = s2.t_id 
		and s1.score > s2.score
	group by
		s1.id 
	having
		count( s1.id )< 2 
	order by
		s1.t_id,
		s1.score 
	) tt 
group by
	t_id

结果：

分析：

1. 查询出所有t1.score>t2.score 的记录

select
		s1.*,s2.*
	from
		student s1
		left join student s2 on s1.t_id = s2.t_id 
		and s1.score > s2.score

2. group by s.id 去重，having 计数取2条

3. group by t_id 分别取各自学科的然后avg取均值

题目三：获取每个科目下前五成绩排名（不允许并列）

select
	* 
from
	(
	select
		s1.*,
		@rownum := @rownum + 1 as num_tmp,
		@incrnum :=
	case
			
			when @rowtotal = s1.score then
			@incrnum 
			when @rowtotal := s1.score then
			@rownum 
		end as rownum 
	from
		student s1
		left join student s2 on s1.t_id = s2.t_id 
		and s1.score > s2.score,
		( select @rownum := 0, @rowtotal := null, @incrnum := 0 ) as it 
	group by
		s1.id 
	order by
		s1.t_id,
		s1.score desc 
	) tt 
group by
	t_id,
	score,
	rownum 
having
	count( rownum )< 5

分析：

1.引入辅助参数

select
	s1.*,
	@rownum := @rownum + 1 as num_tmp,
	@incrnum :=
case
		
		when @rowtotal = s1.score then
		@incrnum 
		when @rowtotal := s1.score then
		@rownum 
	end as rownum 
from
	student s1
	left join student s2 on s1.t_id = s2.t_id 
	and s1.score > s2.score,
	( select @rownum := 0, @rowtotal := null, @incrnum := 0 ) as it

2.去除重复s1.id，分组排序

select
		s1.*,
		@rownum := @rownum + 1 as num_tmp,
		@incrnum :=
	case
			
			when @rowtotal = s1.score then
			@incrnum 
			when @rowtotal := s1.score then
			@rownum 
		end as rownum 
	from
		student s1
		left join student s2 on s1.t_id = s2.t_id 
		and s1.score > s2.score,
		( select @rownum := 0, @rowtotal := null, @incrnum := 0 ) as it 
	group by
		s1.id 
	order by
		s1.t_id,
		s1.score desc

3.group by t_id, score, rownum 然后 having 取前5条不重复的

总结

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MySQL分类排名和分组TOP N实例详解

目录

表结构

题目一：获取每个科目下前五成绩排名（允许并列）

分析：

题目二：获取每个科目下最后两名学生的成绩平均值

分析：

题目三：获取每个科目下前五成绩排名（不允许并列）

分析：

总结

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