MYSQL8查询每个人成绩前三的学科或每个学科前三的成绩

题目

有2张表，用户表，成绩表

• 1、找出每个人成绩前三学科
• 2、找出每个学科前三的成绩

-- 用户表
CREATE TABLE `user_info` (
  `id` int NOT NULL AUTO_INCREMENT,
  `name` varchar(255) COLLATE utf8mb4_general_ci DEFAULT NULL,
  `age` int DEFAULT NULL,
  `email` varchar(255) COLLATE utf8mb4_general_ci DEFAULT NULL,
  PRIMARY KEY (`id`)
) ENGINE=InnoDB AUTO_INCREMENT=4 DEFAULT CHARSET=utf8mb4 COLLATE=utf8mb4_general_ci;

-- 用户数据
INSERT INTO `user_info`(`id`, `name`, `age`, `email`) VALUES (1, 'moss', 33, '[email protected]');
INSERT INTO `user_info`(`id`, `name`, `age`, `email`) VALUES (2, 'jim', 25, '[email protected]');
INSERT INTO `user_info`(`id`, `name`, `age`, `email`) VALUES (3, 'tom', 18, '[email protected]');

-- 成绩表
CREATE TABLE `user_course` (
  `id` int NOT NULL AUTO_INCREMENT,
  `user_id` int DEFAULT NULL,
  `course` varchar(30) COLLATE utf8mb4_general_ci DEFAULT NULL,
  `score` int DEFAULT NULL,
  PRIMARY KEY (`id`)
) ENGINE=InnoDB AUTO_INCREMENT=19 DEFAULT CHARSET=utf8mb4 COLLATE=utf8mb4_general_ci;

-- 成绩数据
INSERT INTO `user_course`(`id`, `user_id`, `course`, `score`) VALUES (1, 1, '语文', 90);
INSERT INTO `user_course`(`id`, `user_id`, `course`, `score`) VALUES (2, 1, '数学', 66);
INSERT INTO `user_course`(`id`, `user_id`, `course`, `score`) VALUES (3, 1, '英语', 80);
INSERT INTO `user_course`(`id`, `user_id`, `course`, `score`) VALUES (4, 1, '物理', 50);
INSERT INTO `user_course`(`id`, `user_id`, `course`, `score`) VALUES (5, 1, '综合', 100);
INSERT INTO `user_course`(`id`, `user_id`, `course`, `score`) VALUES (6, 2, '语文', 88);
INSERT INTO `user_course`(`id`, `user_id`, `course`, `score`) VALUES (7, 2, '数学', 70);
INSERT INTO `user_course`(`id`, `user_id`, `course`, `score`) VALUES (8, 2, '英语', 99);
INSERT INTO `user_course`(`id`, `user_id`, `course`, `score`) VALUES (9, 2, '政治', 70);
INSERT INTO `user_course`(`id`, `user_id`, `course`, `score`) VALUES (10, 2, '艺术', 66);
INSERT INTO `user_course`(`id`, `user_id`, `course`, `score`) VALUES (11, 3, '语文', 68);
INSERT INTO `user_course`(`id`, `user_id`, `course`, `score`) VALUES (12, 3, '画画', 80);
INSERT INTO `user_course`(`id`, `user_id`, `course`, `score`) VALUES (13, 3, '化学', 70);
INSERT INTO `user_course`(`id`, `user_id`, `course`, `score`) VALUES (14, 3, '英语', 98);
INSERT INTO `user_course`(`id`, `user_id`, `course`, `score`) VALUES (15, 4, '语文', 80);
INSERT INTO `user_course`(`id`, `user_id`, `course`, `score`) VALUES (16, 4, '英语', 88);
INSERT INTO `user_course`(`id`, `user_id`, `course`, `score`) VALUES (17, 4, '数学', 90);
INSERT INTO `user_course`(`id`, `user_id`, `course`, `score`) VALUES (18, 4, '历史', 99);

用户数据

成绩数据

1、找出每个人成绩前三学科

解题思路

• 第一步： 成绩表按每人学科成绩排名得出每个人的学科成绩排名

SELECT c1.user_id, c1.course, c1.score, (
		SELECT count(DISTINCT c2.score)
		FROM user_course c2 
		WHERE c1.score < c2.score
			AND c1.user_id = c2.user_id
	) + 1 AS num
FROM user_course c1 
ORDER BY c1.user_id, num

排名结果

• 第二步： 将第一步所得的结果集命名为c3，然后将c3中num小于等于3的查询出来并按学科排序

SELECT c3.user_id, c3.course, c3.score
FROM (
	SELECT c1.user_id, c1.course, c1.score, (
			SELECT count(DISTINCT c2.score)
			FROM user_course c2 
			WHERE c1.score < c2.score
				AND c1.user_id = c2.user_id
		) + 1 AS num
	FROM user_course c1 
	HAVING num <= 3
	ORDER BY c1.user_id, num
) c3 
ORDER BY c3.user_id ASC, c3.score DESC

结果

• 第三步： 对最终查询结果与user_info两表进行左连接查出用户名

select u.name, c4.course, c4.score 
from user_info u 
LEFT JOIN (
	SELECT c3.user_id, c3.course, c3.score
	FROM (
		SELECT c1.user_id, c1.course, c1.score, (
				SELECT count(DISTINCT c2.score)
				FROM user_course c2 
				WHERE c1.score < c2.score
					AND c1.user_id = c2.user_id
			) + 1 AS num
		FROM user_course c1 
		HAVING num <= 3
		ORDER BY c1.user_id, num
	) c3 
	ORDER BY c3.user_id ASC, c3.score DESC
) c4 ON u.id = c4.user_id

结果

2、找出每个学科前三的成绩

思路与上面的一样，先按每个学科的成绩排名，然后筛选出前3的即可

SELECT c3.user_id, c3.course, c3.score
FROM (
	SELECT c1.user_id, c1.course, c1.score, (
			SELECT count(DISTINCT c2.score)
			FROM user_course c2 
			WHERE c1.score <= c2.score
				AND c1.course = c2.course
		) AS num
	FROM user_course c1 
	HAVING num <= 3 
	ORDER BY c1.course, c1.score desc 
) c3 

结果