目录
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
- 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
- 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：
- 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
-
group by s.id 去重，having 计数取2条
-
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
ROW_NUMBER()函数版
必要条件：MySQL版本 8.0+
row_number() OVER ([PARTITION BY COL1] ORDER BY COL2)
表示根据COL1分组，在分组内部根据 COL2排序，而此函数计算的值就表示每组内部排序后的顺序编号（组内连续的唯一的)。
select * from (SELECT
*,
ROW_NUMBER() OVER(PARTITION BY t_id ORDER BY score DESC) top
FROM
student
)t where top<=5< code></=5<>
参考资料：
mysql 每个分类前十名_一文解决所有MySQL分类排名问题
Original: https://blog.csdn.net/Dlgdlgd/article/details/122129327
Author: 奔放的程序猿
Title: MySQL 分类排名（并列、不并列）,分组TOP N,ROW_NUMBER()函数
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