Beispielanalyse des Klassifizierungsrankings und der Gruppierung von TOP N in MySQL

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

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