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经典SQL面试练习50多道题(含答案).docx
以下包含原始数据、问题、答案等，题目有挑战性，认真独立思考之后，不会可看答案、查
资料，需要多复习，直到下次能理清思路，独立完成，才算熟练。
多表查询总结：
1. 确定有哪些表
2. 确定连接条件
3. 确定查询字段
原始数据
一
CREATE TABLE tbl_students (
id INT NOT NULL,
NAME VARCHAR(10) DEFAULT NULL,
sex VARCHAR(10) DEFAULT NULL,
age INT DEFAULT NULL,
PRIMARY KEY (id)
);
INSERT INTO tbl_students (id, NAME, sex, age) VALUES('2','李四','男','21');
INSERT INTO tbl_students (id, NAME, sex, age) VALUES('3','张三','女','17');
INSERT INTO tbl_students (id, NAME, sex, age) VALUES('4','李四','男','12');
INSERT INTO tbl_students (id, NAME, sex, age) VALUES('6','凤姐','女','20');
INSERT INTO tbl_students (id, NAME, sex, age) VALUES('5','凤姐','女','20');
INSERT INTO tbl_students (id, NAME, sex, age) VALUES('7','田七','男','18');
INSERT INTO tbl_students (id, NAME, sex, age) VALUES('1','田七','男','18');
INSERT INTO tbl_students (id, NAME, sex, age) VALUES('8','张三','男','17');
二
CREATE TABLE tbl_score (
id INT NOT NULL,
username VARCHAR(20) DEFAULT NULL,
course VARCHAR(20) DEFAULT NULL,
score INT DEFAULT NULL,
PRIMARY KEY (id)
);
INSERT INTO tbl_score (id, username, course, score) VALUES('1','张三','语文','50');
INSERT INTO tbl_score (id, username, course, score) VALUES('2','张三','数学','80');
INSERT INTO tbl_score (id, username, course, score) VALUES('3','张三','英语','90');
INSERT INTO tbl_score (id, username, course, score) VALUES('4','李四','语文','70');
INSERT INTO tbl_score (id, username, course, score) VALUES('5','李四','数学','80');
INSERT INTO tbl_score (id, username, course, score) VALUES('6','李四','英语','80');
INSERT INTO tbl_score (id, username, course, score) VALUES('7','王五','语文','50');
INSERT INTO tbl_score (id, username, course, score) VALUES('8','王五','英语','70');
INSERT INTO tbl_score (id, username, course, score) VALUES('9','赵六','数学','90');
问题：
1. 删除除学号 id 字段以外，其他字段都相同的冗余数据，只保留一条。
2. 查询参考科目都及格的学员全部信息。
结果图示：
1
2
答案：
1
DELETE FROM tbl_students WHERE id NOT IN(SELECT MID FROM (SELECT MIN(id) MID FROM
tbl_students GROUP BY NAME) t)
2
SELECT * FROM tbl_score WHERE username NOT IN(SELECT username FROM tbl_score WHERE
score<60)
综合练习。建表 sql 语句，如下：
-- 建表
-- 学生表
CREATE TABLE `Student`(
`s_id` VARCHAR(20),
`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`)
);
-- 课程表
CREATE TABLE `Course`(
`c_id` VARCHAR(20),
`c_name` VARCHAR(20) NOT NULL DEFAULT '',
`t_id` VARCHAR(20) NOT NULL,
PRIMARY KEY(`c_id`)
);
-- 教师表
CREATE TABLE `Teacher`(
`t_id` VARCHAR(20),
`t_name` VARCHAR(20) NOT NULL DEFAULT '',
PRIMARY KEY(`t_id`)
);
-- 成绩表
CREATE TABLE `Score`(
`s_id` VARCHAR(20),
`c_id` VARCHAR(20),
`s_score` INT(3),
PRIMARY KEY(`s_id`,`c_id`)
);
-- 插入学生表测试数据
INSERT INTO Student VALUES('01' , '赵雷' , '1990-01-01' , '男');
INSERT INTO Student VALUES('02' , '钱电' , '1990-12-21' , '男');
INSERT INTO Student VALUES('03' , '孙风' , '1990-05-20' , '男');
INSERT INTO Student VALUES('04' , '李云' , '1990-08-06' , '男');
INSERT INTO Student VALUES('05' , '周梅' , '1991-12-01' , '女');
INSERT INTO Student VALUES('06' , '吴兰' , '1992-03-01' , '女');
INSERT INTO Student VALUES('07' , '郑竹' , '1989-07-01' , '女');
INSERT INTO Student VALUES('08' , '王菊' , '1990-01-20' , '女');
-- 课程表测试数据
INSERT INTO Course VALUES('01' , '语文' , '02');
INSERT INTO Course VALUES('02' , '数学' , '01');
INSERT INTO Course VALUES('03' , '英语' , '03');
-- 教师表测试数据
INSERT INTO Teacher VALUES('01' , '张三');
INSERT INTO Teacher VALUES('02' , '李四');
INSERT INTO Teacher VALUES('03' , '王五');
-- 成绩表测试数据
INSERT INTO Score VALUES('01' , '01' , 80);
INSERT INTO Score VALUES('01' , '02' , 90);
INSERT INTO Score VALUES('01' , '03' , 99);
INSERT INTO Score VALUES('02' , '01' , 70);
INSERT INTO Score VALUES('02' , '02' , 60);
INSERT INTO Score VALUES('02' , '03' , 80);
INSERT INTO Score VALUES('03' , '01' , 80);
INSERT INTO Score VALUES('03' , '02' , 80);
INSERT INTO Score VALUES('03' , '03' , 80);
INSERT INTO Score VALUES('04' , '01' , 50);
INSERT INTO Score VALUES('04' , '02' , 30);
INSERT INTO Score VALUES('04' , '03' , 20);
INSERT INTO Score VALUES('05' , '01' , 76);
INSERT INTO Score VALUES('05' , '02' , 87);
INSERT INTO Score VALUES('06' , '01' , 31);
INSERT INTO Score VALUES('06' , '03' , 34);
INSERT INTO Score VALUES('07' , '02' , 89);
INSERT INTO Score VALUES('07' , '03' , 98);
问题与答案：（题目多，答案未必对，请知悉。）
-- 1、查询"01"课程比"02"课程成绩高的学生的信息及课程分数
SELECT a.*,b.s_score AS 01_score,c.s_score AS 02_score FROM
student a
JOIN score b ON a.s_id=b.s_id AND b.c_id='01'
LEFT JOIN score c ON a.s_id=c.s_id AND c.c_id='02' OR c.c_id = NULL WHERE b.s_score>c.s_score
-- 也可以这样写
SELECT a.*,b.s_score AS 01_score,c.s_score AS 02_score 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'
AND b.s_score>c.s_score
-- 2、查询"01"课程比"02"课程成绩低的学生的信息及课程分数
SELECT a.*,b.s_score AS 01_score,c.s_score AS 02_score FROM
student a LEFT JOIN score b ON a.s_id=b.s_id AND b.c_id='01' OR b.c_id=NULL
JOIN score c ON a.s_id=c.s_id AND c.c_id='02' WHERE b.s_score=60;
-- 4、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 b.s_id,b.s_name,ROUND(AVG(a.s_score),2) AS avg_score FROM
student b
LEFT JOIN score a ON b.s_id = a.s_id
GROUP BY b.s_id,b.s_name HAVING avg_score <60
UNION
SELECT a.s_id,a.s_name,0 AS avg_score FROM
student a
WHERE a.s_id NOT IN (
SELECT DISTINCT s_id FROM score);
-- 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 a.s_id,a.s_name,COUNT(b.c_id) AS sum_course,SUM(b.s_score) AS sum_score FROM
student a
LEFT JOIN score b ON a.s_id=b.s_id
GROUP BY a.s_id,a.s_name;
-- 6、查询"李"姓老师的数量
SELECT COUNT(t_id) FROM teacher WHERE t_name LIKE '李%';
-- 7、查询学过"张三"老师授课的同学的信息
SELECT a.* FROM
student a
JOIN score b ON a.s_id=b.s_id WHERE b.c_id IN(
SELECT c_id FROM course WHERE t_id =(
SELECT t_id FROM teacher WHERE t_name = '张三'));
-- 8、查询没学过"张三"老师授课的同学的信息
SELECT * FROM
student c
WHERE c.s_id NOT IN(
SELECT a.s_id FROM student a JOIN score b ON a.s_id=b.s_id WHERE b.c_id IN(
SELECT a.c_id FROM course a JOIN teacher b ON a.t_id = b.t_id WHERE t_name ='张三
'));
-- 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、查询没有学全所有课程的同学的信息
-- @wendiepei 的写法
SELECT s.* FROM student s
LEFT JOIN Score s1 ON s1.s_id=s.s_id
GROUP BY s.s_id HAVING COUNT(s1.c_id)<(SELECT COUNT(*) FROM course)
-- @k1051785839 的写法
SELECT *
FROM student
WHERE s_id NOT IN(
SELECT s_id FROM score t1
GROUP BY s_id HAVING COUNT(*) =(SELECT COUNT(DISTINCT c_id)
FROM course))
-- 12、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')
);
-- 13、查询和"01"号的同学学习的课程完全相同的其他同学的信息
-- @ouyang_1993 的写法
SELECT
Student.*
FROM
Student
WHERE
s_id IN (SELECT s_id FROM Score GROUP BY s_id HAVING COUNT(s_id) = (
#下面的语句是找到'01'同学学习的课程数
SELECT COUNT(c_id) FROM Score WHERE s_id = '01'
)
)
AND s_id NOT IN (
#下面的语句是找到学过‘01’同学没学过的课程，有哪些同学。并排除他们
SELECT s_id FROM Score
WHERE c_id IN(
#下面的语句是找到‘01’同学没学过的课程
SELECT DISTINCT c_id FROM Score
WHERE c_id NOT IN (
#下面的语句是找出‘01’同学学习的课程
SELECT c_id FROM Score WHERE s_id = '01'
)
) GROUP BY s_id
) #下面的条件是排除 01 同学
AND s_id NOT IN ('01')
-- @k1051785839 的写法
SELECT
t3.*
FROM
(
SELECT
s_id,
GROUP_CONCAT(c_id ORDER BY c_id) group1
FROM
score
WHERE
s_id <> '01'
GROUP BY
s_id
) t1
INNER JOIN (
SELECT
GROUP_CONCAT(c_id ORDER BY c_id) group2
FROM
score
WHERE
s_id = '01'
GROUP BY
s_id
) t2 ON t1.group1 = t2.group2
INNER JOIN student t3 ON t1.s_id = t3.s_id
-- 14、查询没学过"张三"老师讲授的任一门课程的学生姓名
SELECT a.s_name FROM student a WHERE a.s_id NOT IN (
SELECT s_id FROM score WHERE c_id =
(SELECT c_id FROM course WHERE t_id =(
SELECT t_id FROM teacher WHERE t_name = '张三')));
-- 15、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.s_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 s_score<60 GROUP BY
GROUP BY a.s_id,a.s_name
s_id HAVING COUNT(1)>=2)
-- 16、检索"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.s_score FROM
student a,score b
WHERE a.s_id = b.s_id AND b.c_id='01' AND b.s_score<60 ORDER BY b.s_score DESC;
-- 17、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,(SELECT s_score FROM score WHERE s_id=a.s_id AND c_id='01') AS 语文,
(SELECT s_score FROM score WHERE s_id=a.s_id AND c_id='02') AS 数学,
(SELECT s_score FROM score WHERE s_id=a.s_id AND c_id='03') AS 英语,
ROUND(AVG(s_score),2) AS 平均分 FROM score a GROUP BY a.s_id ORDER BY 平均分 DESC;
-- @喝完这杯还有一箱的写法
SELECT a.s_id,MAX(CASE a.c_id WHEN '01' THEN a.s_score END ) 语文,
MAX(CASE a.c_id WHEN '02' THEN a.s_score END ) 数学,
MAX(CASE a.c_id WHEN '03' THEN a.s_score END ) 英语,
AVG(a.s_score),b.s_name FROM Score a JOIN Student b ON a.s_id=b.s_id GROUP BY a.s_id
ORDER BY 5 DESC
-- 18.查询各科成绩最高分、最低分和平均分：以如下形式显示：课程 ID，课程 name，最高
分，最低分，平均分，及格率，中等率，优良率，优秀率
-- 及格为>=60，中等为：70-80，优良为：80-90，优秀为：>=90
SELECT a.c_id,b.c_name,MAX(s_score),MIN(s_score),ROUND(AVG(s_score),2),
ROUND(100*(SUM(CASE WHEN a.s_score>=60 THEN 1 ELSE 0 END)/SUM(CASE WHEN a.s_score
THEN 1 ELSE 0 END)),2) AS 及格率,
ROUND(100*(SUM(CASE WHEN a.s_score>=70 AND a.s_score<=80
END)/SUM(CASE WHEN a.s_score THEN 1 ELSE 0 END)),2) AS 中等率,
ROUND(100*(SUM(CASE WHEN a.s_score>=80 AND a.s_score<=90
END)/SUM(CASE WHEN a.s_score THEN 1 ELSE 0 END)),2) AS 优良率,
ROUND(100*(SUM(CASE WHEN a.s_score>=90 THEN 1 ELSE 0 END)/SUM(CASE WHEN a.s_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
THEN 1
THEN 1
0
0
ELSE
ELSE
-- 19、按各科成绩进行排序，并显示排名
-- mysql 没有 rank 函数
SELECT a.s_id,a.c_id,
@i:=@i +1 AS i 保留排名,
@k:=(CASE WHEN @score=a.s_score THEN @k ELSE @i END) AS rank 不保留排名,
@score:=a.s_score AS score
FROM (
SELECT s_id,c_id,s_score FROM score GROUP BY s_id,c_id,s_score ORDER BY s_score
DESC
)a,(SELECT @k:=0,@i:=0,@score:=0)s
--@k1051785839 的写法
(SELECT * FROM (SELECT
t1.c_id,
t1.s_score,
(SELECT COUNT(DISTINCT t2.s_score) FROM score t2 WHERE t2.s_score>=t1.s_score AND
t2.c_id='01') rank
FROM score t1 WHERE t1.c_id='01'
ORDER BY t1.s_score DESC) t1)
UNION
(SELECT * FROM (SELECT
t1.c_id,
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