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MySQL中分类排名和分组TOP N的示例分析

发表于:2024-10-02 作者:千家信息网编辑
千家信息网最后更新 2024年10月02日,这篇文章主要介绍MySQL中分类排名和分组TOP N的示例分析,文中介绍的非常详细,具有一定的参考价值,感兴趣的小伙伴们一定要看完!表结构学生表如下:CREATE TABLE `t_student`
千家信息网最后更新 2024年10月02日MySQL中分类排名和分组TOP N的示例分析

这篇文章主要介绍MySQL中分类排名和分组TOP N的示例分析,文中介绍的非常详细,具有一定的参考价值,感兴趣的小伙伴们一定要看完!

    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.idHAVING        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.idHAVING        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

    以上是"MySQL中分类排名和分组TOP N的示例分析"这篇文章的所有内容,感谢各位的阅读!希望分享的内容对大家有帮助,更多相关知识,欢迎关注行业资讯频道!

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