【Mysql学习之旅-2】经典sql面试题及答案分析
前言
在学习了Mysql的基础知识后，我们用一套sql练习题来实战一下。
关于练习所需要的数据，让我们继续玩坏倚天屠龙的江湖。
1、学生表 student（s_id：学生id，s_name：学生姓名，s_birth：学生生日，s_sex：学生性别）：
2、教师表teacher（t_id：教师id，t_name：教师姓名)
3、课程表 course（c_id：课程id，c_name：课程名称，t_id：教师id）：
4、成绩表 score（s_id：学生id，c_id：课程id，score：分数）
初始化数据的sql附在文章末尾
题目开始，先上几道硬菜，磕到牙齿的可以移步进阶题目和基础题目区压压惊。
挑战
1、查询各科成绩前三名的记录
--方法1：
SELECT s_name,a.s_id,a.c_id,a.score
FROM score a
LEFT JOIN score b
ON a.c_id = b.c_id AND a.score<b.score
JOIN student
ONa.s_id=student.s_id
GROUP BY a.s_id,a.c_id,a.score HAVING COUNT(b.s_id)<3
ORDER BY a.c_id,a.score DESC--方法2：
SELECT a.s_id,a.c_id,a.score
FROM score a
WHERE (SELECT COUNT(1) FROM score b WHERE b.c_id=a.c_id AND b.score>=a.score)<=2 ORDER BY a.c_id
简评：非常难以理解的一道题目，即使想透彻了也不容易记住，堪称劝退型难题。面试中，偶尔会考察。
难度：⭐⭐⭐⭐⭐
易考率：⭐⭐
思路：
1、SELECT * FROM score a JOIN score b ON a.c_id=b.c_id WHERE a.c_id=‘03’ 这一句，形成了03课程的所有学员的两两组合（带上03，是为了简化数据，便于分析）
2、SELECT * FROM score a JOIN score b ON a.c_id=b.c_id WHERE a.c_id=‘03’ AND a.score<b.score又筛选出每个学员03课程比其他人低的情况（某个学员分越低，记录越多，没记录说明是最高分）
3、将这些记录按s_id分组，筛选出记录数小于3的（说明最多有两个人比自己分数高），即是前3名
2、查询“01”课程比“02”课程成绩高的所有学生的学号
SELECT *
FROM (SELECT s_id,score,c_id FROM score WHERE c_id='01') a
JOIN(SELECT s_id,score,c_id FROM score WHERE c_id='02') b
ON a.s_id=b.s_id
WHERE a.score>b.score
简评：不容易想到，但答案相对第1题来说，就好理解多了。
难度：⭐⭐⭐⭐
频率：⭐⭐
思路：通过SELECT s_id,score,c_id FROM score WHERE c_id='01’和SELECT s_id,score,c_id FROM score WHERE c_id='02’的两个临时表联查，得到每个学员的01、02课程成绩情况：
接下来，a.score>b.score条件直接筛选就可以了。
3、查询和"04"号的同学学习的课程完全相同的其他同学的信息
SELECT score.s_id,COUNT(DISTINCT c_id)
FROM student JOIN score ON student.s_id=score.s_id
GROUP BY score.s_id
HAVING COUNT(DISTINCT c_id)= (SELECT COUNT(c_id) FROM score WHERE s_id='04')
AND s_id
NOT IN-- 筛选出学过（04号同学未学课程）的学生，通过not in排除掉(SELECT s_id FROM score WHERE c_id IN(SELECT DISTINCT(c_id) FROM course WHERE c_id NOT IN(SELECT c_id FROM score WHERE s_id='04')))
AND score.s_id!='04'
简评：难在思路的迂回性。面试中，很多有难度的sql也大多难在直接去考虑，难以获得答案。
难度：⭐⭐⭐⭐
频率：⭐⭐
思路：通常思路，我们先获取04号同学的课程，再遍历其他同学，以此筛选。这种方式适用于编程，单纯的sql实现不了，我们应该从下面这个思路去考虑------>我和04号同学学习的课程数目一样多，且04号同学没学过的我也没学过，那么我不就是和04号同学学习的课程一样吗？
4、按平均成绩从高到低显示所有学生的所有课程的成绩以及平均成绩
SELECT a.s_id,b.s_name,MAX(CASE a.c_id WHEN '01' THEN a.score END ) 内功, MAX(CASE a.c_id WHEN '02' THEN a.score END ) 剑法, MAX(CASE a.c_id WHEN '03' THEN a.score END ) 拳法, AVG(a.score)
FROM score a
JOIN student b ON a.s_id=b.s_id
GROUP BY a.s_id ORDER BY AVG(a.score) DESC
简评：思路不难，关键在于理解透彻case when语法。
难度：⭐⭐⭐
频率：⭐⭐
SELECT a.c_id,b.c_name,MAX(score),MIN(score),ROUND(AVG(score),2),ROUND(100*(SUM(case when a.score>=60 then 1 else 0 end)/SUM(case when a.score then 1 else 0 end)),2) as 及格率,ROUND(100*(SUM(case when a.score>=70 and a.score<=80 then 1 else 0 end)/SUM(case when a.score then 1 else 0 end)),2) as 中等率,ROUND(100*(SUM(case when a.score>=80 and a.score<=90 then 1 else 0 end)/SUM(case when a.score then 1 else 0 end)),2) as 优良率,ROUND(100*(SUM(case when a.score>=90 then 1 else 0 end)/SUM(case when a.score then 1 else 0 end)),2) as 优秀率
FROM score a
LEFT JOIN course b
ON a.c_id = b.c_id GROUP BY a.c_id,b.c_name
简评：在case when语法的基础上，进一步掌握sql语句中进行逻辑运算的技巧
难度：★★★☆
频率：★★★
5、按各科成绩进行排序，并显示排名
SELECT a.s_id,a.c_id,@i:=@i + 1 AS 排名,@j:=(CASE WHEN @score=a.score THEN @j ELSE @i END) AS 并排排名,@score:=a.score AS score
FROM (SELECT s_id,c_id,score FROM score GROUP BY s_id,c_id ORDER BY score DESC) a,(SELECT @i:= 0,@j:= 0,@score:= 0) s
简评：Mysql中没有rank函数，只能使用伪列的概念去实现排名算法。
难度：★★★★
频率：★★
思路：
1、SELECT s_id,c_id,score FROM score GROUP BY s_id,c_id ORDER BY score DESC先进行了排序
2、(SELECT @i:=0,@j:=0,@score:=0) s对@i、@j、@score进行了初始化，初始值都为0
3、@i:=@i + 1，遍历每一行，@i逐行自增
4、@j:=(CASE WHEN @score=a.score THEN @j ELSE @i END)，遍历每一行，逻辑判断，@score与上一行的score相等则@j=@i，否则则自增一次
6、查询学生的总成绩并进行排名
SELECT a.s_id,@i:=@i+1 as i,@j:=(CASE WHEN @score=a.sum_score THEN @j ELSE @i END) AS rank,@score:=a.sum_score AS score
FROM (SELECT s_id,SUM(score) AS sum_score FROM score GROUP BY s_id ORDER BY sum_score DESC)a,(SELECT @i:=0,@j:=0,@score:=0) s
简评：与上一题基本差不多，不同的是a这个临时表的数据内容而已。
难度：★★★★
频率：★★
7、查询所有课程的成绩第2名到第3名的学生信息及该课程成绩
SELECT student.*,c.rank,c.score,c.c_id
FROM (SELECT a.s_id,a.score,a.c_id,@i:=@i+1 as rank from score a,(SELECT @i:=0)s WHERE a.c_id='01' ORDER BY a.score DESC) c
LEFT JOIN student ON c.s_id=student.s_id
WHERE rank BETWEEN 2 AND 3
UNION ALL
SELECT student.*,c.rank,c.score,c.c_id
FROM (SELECT a.s_id,a.score,a.c_id,@j:=@j+1 as rank from score a,(SELECT @j:=0)s WHERE a.c_id='02' ORDER BY a.score DESC) c
LEFT JOIN student ON c.s_id=student.s_id
WHERE rank BETWEEN 2 AND 3
UNION ALL
SELECT student.*,c.rank,c.score,c.c_id
FROM (SELECT a.s_id,a.score,a.c_id,@k:=@k+1 as rank from score a,(SELECT @k:=0)s WHERE a.c_id='03' ORDER BY a.score DESC) c
LEFT JOIN student ON c.s_id=student.s_id
WHERE rank BETWEEN 2 AND 3
简评：sql很长，不要被吓到，其实是3个相同的部分（where条件不同）通过UNION ALL合并而已。另外，筛选2、3名也是在排名的基础上进一步筛选。
难度：★★★★☆
频率：★★
思路：
1、先将01课程的所有人的分数排序，再进一步根据rank BETWEEN 2 AND 3筛选出2、3名
2、使用同样方法，筛选02、03课程数据，使用UNION ALL合并查询结果
8、统计各科成绩各分数段人数：课程编号、课程名称、[80-100]，[60-80]，[0-60]及所占百分比
SELECT DISTINCT e.c_name,a.c_id,b.`80-100`,b.百分比,c.`60-80`,c.百分比,d.`0-60`,d.百分比
FROM score a
LEFT JOIN (SELECT c_id,SUM(CASE WHEN score >80 AND score <=100 THEN 1 ELSE 0 END) AS `80-100`,ROUND(100*(SUM(CASE WHEN score >80 AND score <=100 THEN 1 ELSE 0 end)/count(*)),2)AS 百分比 FROM score GROUP BY c_id)b ON a.c_id=b.c_id
LEFT JOIN (SELECT c_id,SUM(CASE WHEN score >60 AND score <=80 THEN 1 ELSE 0 END) AS `60-80`,ROUND(100*(SUM(CASE WHEN score >60 AND score <=80 THEN 1 ELSE 0 end)/count(*)),2) AS 百分比 FROM score GROUP BY c_id)c ON a.c_id=c.c_id
LEFT JOIN (SELECT c_id,SUM(CASE WHEN score >=0 AND score <=60 THEN 1 ELSE 0 END) AS `0-60`,ROUND(100*(SUM(CASE WHEN score >=0 AND score <=60 THEN 1 ELSE 0 END)/count(*)),2) AS 百分比 FROM score GROUP BY c_id)d ON a.c_id=d.c_id
LEFT JOIN course e ON a.c_id = e.c_id
简评：
CASE WHEN和sql语句内进行逻辑运算的又一次应用。
难度：★★★★
频率：★★
思路：SELECT c_id,SUM(CASE WHEN score >80 AND score <=100 THEN 1 ELSE 0 END) AS80-100, ROUND(100*(SUM(CASE WHEN score >80 AND score <=100 THEN 1 ELSE 0 END)/count(*)),2) AS 百分比 FROM score GROUP BY c_id得到所有课程80-100分成绩的统计临时表：
然后再用同样的方式，得到0-60、60-80的统计临时表，三表联查。
9、查询学生平均成绩及其名次
SELECT a.s_id,@i:=@i+1 as '不保留空缺排名',@k:=(CASE WHEN @avg_score=a.avg_s THEN @k ELSE @i END) AS '保留空缺排名',@avg_score:=avg_s AS '平均分'
FROM (SELECT s_id,ROUND(AVG(score),2) AS avg_s FROM score GROUP BY s_id ORDER BY avg_s DESC)a,(SELECT @avg_score:=0,@i:=0,@k:=0)b;
简评：与第5题类似，没什么特别需要说的。
难度：★★★★
频率：★★
10、查询各学生的年龄
SELECT s_birth,(DATE_FORMAT(NOW(),'%Y')-DATE_FORMAT(s_birth,'%Y') - (CASE WHEN DATE_FORMAT(NOW(),'%m%d')>DATE_FORMAT(s_birth,'%m%d') THEN 0 ELSE 1 END)) AS age
FROM student;
简评：本身的实现思路并不难，
DATE_FORMAT掌握其用法即可。
11、分别查询本周、下周、本月、下月过生日的学生
--本周
SELECT * FROM student WHERE WEEK(DATE_FORMAT(NOW(),'%Y%m%d'))=WEEK(s_birth)
--下周
SELECT * FROM student WHERE WEEK(DATE_FORMAT(NOW(),'%Y%m%d'))+1 = WEEK(s_birth)
--本月
SELECT * FROM student WHERE MONTH(DATE_FORMAT(NOW(),'%Y%m%d')) =MONTH(s_birth)
--下月
SELECT * FROM student WHERE MONTH(DATE_FORMAT(NOW(),'%Y%m%d'))+1 =MONTH(s_birth)
进阶
1、查询及格学生（每门课程的分数>=60）的学生姓名
SELECT s_name FROM student
WHERE s_id NOT IN(SELECT DISTINCT(s_id) FROM score WHERE score<60)
或者：
SELECT s_name FROM student a JOIN score b ON a.s_id=b.s_id
GROUP BY b.s_id HAVING MIN(score)>60
2、查询平均成绩小于60分的同学的学生编号和学生姓名和平均成绩 (包括有成绩的和无成绩的)
SELECT s_name,ROUND(avg(score),2) AS avg
FROM student
JOIN score ON student.s_id=score.s_id
GROUP BY score.s_id HAVING avg<60
UNION ALL
SELECT s_name,0 AS avg
FROM student
WHERE s_id NOT IN(SELECT DISTINCT(s_id) FROM score)-- 方法2
SELECT s_name,IFNULL(ROUND(avg(score),2),0) AS avg
FROM student
LEFT JOIN score ON student.s_id=score.s_id
GROUP BY score.s_id HAVING avg<60 OR avg is NULL
3、查询学过"张三丰"老师授课的同学的信息
SELECT *
FROM score JOIN student ON score.s_id=student.s_id
WHERE c_id IN(SELECT c_id FROM course c JOIN teacher t ON c.t_id=t.t_id WHERE t_name='张三丰')
4、查询没学过"张三丰"老师授课的同学的信息
SELECT *
FROM student
WHERE s_id NOT IN(SELECT s_id FROM score WHERE c_id
IN(SELECT c_id FROM course c JOIN teacher t ON c.t_id=t.t_id WHERE t_name='张三丰'))
5、查询没有学全所有课程的同学的信息
SELECT s_name,COUNT(c_id) AS count
FROM student
LEFT JOIN score ON student.s_id=score.s_id
GROUP BY student.s_id HAVING count<(SELECT COUNT(1) FROM course);
6、查询至少有一门课与学号为"01"的同学所学相同的同学的信息
SELECT * FROM student
WHERE s_id IN(SELECT DISTINCT a.s_id FROM score a WHERE a.c_id IN(SELECT a.c_id FROM score a WHERE a.s_id='01'));
7、查询两门及其以上不及格课程的同学的学号，姓名及其平均成绩
SELECT a.s_id,a.s_name,ROUND(AVG(b.score))
FROM student a
LEFT JOIN score b ON a.s_id = b.s_id
WHERE a.s_id IN(SELECT s_id from score WHERE score<60 GROUP BY s_id HAVING count(1)>=2)
GROUP BY a.s_id,a.s_name
8、查询所有学生的课程及分数情况；
SELECT a.s_id,a.s_name,SUM(CASE c.c_name WHEN '内功' THEN b.score ELSE 0 END) AS '内功',SUM(CASE c.c_name WHEN '剑法' THEN b.score ELSE 0 END) AS '剑法',SUM(CASE c.c_name WHEN '拳法' THEN b.score ELSE 0 END) AS '拳法',SUM(b.score) as '总分'
FROM student a
LEFT JOIN score b ON a.s_id = b.s_id
LEFT JOIN course c ON b.c_id = c.c_id
GROUP BY a.s_id,a.s_name
9、查询选修"张三丰"老师所授课程的学生中，成绩最高的学生信息及其成绩
SELECT s_id,MAX(score)
FROM score
WHERE c_id
IN(SELECT c_id FROM teacher JOIN course ON teacher.t_id=course.t_id WHERE t_name='张三丰')
10、查询不同课程成绩相同的学生的学生编号、课程编号、学生成绩
SELECT * FROM score a,score b where a.c_id != b.c_id and a.score = b.score
11、统计每门课程的学生选修人数（超过5人的课程才统计）。要求输出课程号和选修人数，查询结果按人数降序排列，若人数相同，按课程号升序排列
SELECT c_id,COUNT(*) AS total
FROM score
GROUP BY c_id HAVING total>4 ORDER BY total DESC,c_id ASC
12、查询选修了全部课程的学生信息
SELECT * FROM student WHERE s_id
IN(SELECT s_id FROM score GROUP BY s_id HAVING COUNT(*)=(SELECT COUNT(*) FROM course))
基础
1、查询不及格的课程
SELECT a.s_id,a.c_id,b.c_name,a.score
FROM score a
LEFT JOIN course b ON a.c_id = b.c_id
WHERE a.score<60
2、查询课程编号为01且课程成绩在80分以上的学生的学号和姓名
SELECT a.s_id,b.s_name
FROM score a
LEFT JOIN student b ON a.s_id = b.s_id
WHERE a.c_id = '01' AND a.score>80
3、查询每个同学的姓名、选课数、总成绩
SELECT s_name,COUNT(c_id),SUM(score)
FROM student a
LEFT JOIN score b ON a.s_id=b.s_id
GROUP BY b.s_id
4、查询平均成绩大于等于60分的同学的学生编号和学生姓名和平均成绩
SELECT s_name,ROUND(avg(score),2) AS avg
FROM student
JOIN score ON student.s_id=score.s_id
GROUP BY score.s_id HAVING avg>60
5、查询所有同学的学生编号、学生姓名、选课总数、所有课程的总成绩
SELECT student.s_name,COUNT(c_id),IFNULL(SUM(score),0) AS sum FROM student LEFT JOIN score ON student.s_id=score.s_id GROUP BY student.s_id ORDER BY sum DESC
6、查询"张"姓老师的数量
SELECT COUNT(t_id) FROM teacher WHERE t_name LIKE '张%'
7、查询"01"课程分数小于60，按分数降序排列的学生信息
SELECT a.*,b.c_id,b.score
FROM student a,score b
WHERE a.s_id=b.s_id
ANDb.c_id='01' AND b.score<60
ORDER BYb.score DESC;
8、查询任何一门课程成绩在70分以上的姓名、课程名称和分数
SELECT a.s_name,b.c_name,c.score
FROM course b
LEFT JOIN score c ON b.c_id = c.c_id
LEFT JOIN student a ON a.s_id=c.s_id
WHERE c.score>=70
9、查询学过编号为"01"并且也学过编号为"02"的课程的同学的信息
SELECT a.* FROM student a,score b,score c
WHERE a.s_id = b.s_id AND a.s_id = c.s_id AND b.c_id='01' AND c.c_id='02';
10、查询学过编号为"01"但是没有学过编号为"02"的课程的同学的信息
SELECT a.* FROM student a
WHERE a.s_id
IN (SELECT s_id FROM score WHERE c_id='01' )
AND a.s_id NOT IN(SELECT s_id FROM score WHERE c_id='02')
11、查询男生、女生人数
SELECT s_sex,COUNT(s_sex) FROM student GROUP BY s_sex
12、查询名字中含有"圆"字的学生信息
SELECT * FROM student WHERE s_name LIKE '%圆%';
13、查询同名同性学生名单，并统计同名人数
SELECT a.s_name,a.s_sex,count(*)
FROM student a JOIN student b
ON a.s_id !=b.s_id and a.s_name = b.s_name
AND a.s_sex = b.s_sex
GROUP BY a.s_name,a.s_sex
14、查询1990年出生的学生名单
SELECT s_name FROM student WHERE s_birth LIKE '1337%'
15、查询每门课程的平均成绩，结果按平均成绩降序排列，平均成绩相同时，按课程编号升序排列
SELECT c_id,ROUND(AVG(score),2) AS avg_score
FROM score
GROUP BY c_id ORDER BY avg_score DESC,c_id ASC
16、查询平均成绩大于等于85的所有学生的学号、姓名和平均成绩
SELECT a.s_id,b.s_name,ROUND(avg(a.score),2) AS avg
FROM score a
LEFT JOIN student b on a.s_id=b.s_id
GROUP BY s_id HAVING avg>=85
17、查询课程名称为"内功"，且分数低于60的学生姓名和分数
SELECT a.s_name,b.score
FROM score b
JOIN student a ON a.s_id=b.s_id
WHERE b.c_id=(SELECT c_id FROM course WHERE c_name ='内功')
AND b.score<60
18、求每门课程的学生人数
SELECT c_id,count(*) FROM score GROUP BY c_id;
19、检索至少选修两门课程的学生学号
SELECT s_id,count(*) AS count FROM score GROUP BY s_id HAVING count>=2
数据初始化
-- ----------------------------
-- Table structure for course
-- ----------------------------
DROP TABLE IF EXISTS `student`;
CREATE TABLE `student` (`s_id` varchar(20) NOT NULL,`s_name` varchar(20) NOT NULL DEFAULT '',`s_birth` varchar(20) NOT NULL DEFAULT '',`s_sex` varchar(10) NOT NULL DEFAULT '',PRIMARY KEY (`s_id`)
) ENGINE=InnoDB DEFAULT CHARSET=utf8;-- ----------------------------
-- Records of student
-- ----------------------------
INSERT INTO `student` VALUES ('01', '张无忌', '1337-06-17', '男');
INSERT INTO `student` VALUES ('02', '宋青书', '1333-12-21', '男');
INSERT INTO `student` VALUES ('03', '丁敏君', '1336-06-21', '女');
INSERT INTO `student` VALUES ('04', '周芷若', '1340-09-26', '女');
INSERT INTO `student` VALUES ('05', '圆真', '1320-12-01', '男');
INSERT INTO `student` VALUES ('06', '圆觉', '1316-03-01', '男');
INSERT INTO `student` VALUES ('07', '殷梨亭', '1310-07-01', '男');
INSERT INTO `student` VALUES ('08', '纪晓芙', '1310-01-20', '女');-- ----------------------------
-- Table structure for course
-- ----------------------------
DROP TABLE IF EXISTS `course`;
CREATE TABLE `course` (`c_id` varchar(20) NOT NULL,`c_name` varchar(20) NOT NULL DEFAULT '',`t_id` varchar(20) NOT NULL,PRIMARY KEY (`c_id`)
) ENGINE=InnoDB DEFAULT CHARSET=utf8;-- ----------------------------
-- Records of course
-- ----------------------------
INSERT INTO `course` VALUES ('01', '内功', '03');
INSERT INTO `course` VALUES ('02', '剑法', '02');
INSERT INTO `course` VALUES ('03', '拳法', '01');-- ----------------------------
-- Table structure for teacher
-- ----------------------------
DROP TABLE IF EXISTS `teacher`;
CREATE TABLE `teacher` (`t_id` varchar(20) NOT NULL,`t_name` varchar(20) NOT NULL DEFAULT '',PRIMARY KEY (`t_id`)
) ENGINE=InnoDB DEFAULT CHARSET=utf8;-- ----------------------------
-- Records of teacher
-- ----------------------------
INSERT INTO `teacher` VALUES ('01', '张三丰');
INSERT INTO `teacher` VALUES ('02', '灭绝师太');
INSERT INTO `teacher` VALUES ('03', '空见');-- ----------------------------
-- Table structure for score
-- ----------------------------
DROP TABLE IF EXISTS `score`;
CREATE TABLE `score` (`s_id` varchar(20) NOT NULL,`c_id` varchar(20) NOT NULL,`score` int(3) DEFAULT NULL,PRIMARY KEY (`s_id`,`c_id`)
) ENGINE=InnoDB DEFAULT CHARSET=utf8;-- ----------------------------
-- Records of score
-- ----------------------------
INSERT INTO `score` VALUES ('01', '01', '99');
INSERT INTO `score` VALUES ('01', '02', '96');
INSERT INTO `score` VALUES ('01', '03', '90');
INSERT INTO `score` VALUES ('02', '01', '59');
INSERT INTO `score` VALUES ('02', '02', '75');
INSERT INTO `score` VALUES ('02', '03', '68');
INSERT INTO `score` VALUES ('03', '01', '55');
INSERT INTO `score` VALUES ('03', '02', '50');
INSERT INTO `score` VALUES ('03', '03', '60');
INSERT INTO `score` VALUES ('04', '01', '77');
INSERT INTO `score` VALUES ('04', '02', '88');
INSERT INTO `score` VALUES ('05', '01', '88');
INSERT INTO `score` VALUES ('05', '02', '90');
INSERT INTO `score` VALUES ('06', '01', '86');
INSERT INTO `score` VALUES ('06', '03', '99');
INSERT INTO `score` VALUES ('07', '02', '58');
INSERT INTO `score` VALUES ('07', '03', '98');
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