MySQL分类排名和分组TOP N实例详解
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2022-06-10 11:33:38
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:总结表结构学生表如下:creat...
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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