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经典SQL 50题 | 附带优化方案（更新中）

news 2025/10/11 7:34:00

原题目题目链接：
SQL面试必会50题 - 小番茄的文章 - 知乎

ps：本文仅包含50题里的重点题目。

1.案例数据建立参考

表名和字段

1.学生表
Student(s_id,s_name,s_birth,s_sex) –学生编号,学生姓名, 出生年月,学生性别

2.课程表
Course(c_id,c_name,t_id) – –课程编号, 课程名称, 教师编号

3.教师表
Teacher(t_id,t_name) –教师编号,教师姓名

4.成绩表
Score(s_id,c_id,s_score) –学生编号,课程编号,分数

测试数据

--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);

2.题目

7.查询学过编号为“01”的课程并且也学过编号为“02”的课程的学生的学号、姓名（重点）
SQL1:

SELECT st.s_id, st.s_name FROM student st 
WHERE EXISTS(SELECT 1 FROM score sc01 INNER JOIN score sc02 ON sc01.s_id = sc02.s_id AND sc01.c_id = '01' AND sc02.c_id = '02' WHERE st.s_id = sc01.s_id
);

SQL 2： ☑️优解

select st.s_id,st.s_name from student st 
where EXISTS(select 1 from score sc01 where sc01.s_id = st.s_id and sc01.c_id ='01') 
and 
EXISTS (select 1 from score sc02 where sc02.s_id = st.s_id and sc02.c_id ='02')

explain 执行计划分析:

SQL1:
SQL2:

原因：

SQL2 避免不必要的自连接操作
更好的索引利用和短路优化
执行计划更简洁高效

10.查询没有学全所有课的学生的学号、姓名(重点)

select st.s_id from student st left join score sc on st.s_id = sc.s_id GROUP BY st.s_id having count(st.s_id) < (select count(c_id) from course)

注意空数据，因为有可能有学生一门课都没有选。

11.查询至少有一门课与学号为01的学生所学课程相同的学生学号和姓名（重点）

SQL：

select  st.s_id,st.s_name from student st where st.s_id in  (select sc.s_id from score sc where sc.c_id in (select sc2.c_id from score sc2 where sc2.s_id = '01'))

sql思路：

永远永远先过滤~
不要无脑联表查询

1）先查询出01号学生所学的课程号：

select sc2.c_id from score sc2 where sc2.s_id = '01'

2）使用in查询至少有一门课与学号为01的学生所学课程相同的学生id：

select sc.s_id from score sc where sc.c_id in (select sc2.c_id from score sc2 where sc2.s_id = '01')

3）查询学生表符合条件的行数据，完整的SQL：

select st.s_id,st.s_name from student st inner join  (select distinct sc.s_id from score sc where sc.c_id in (select sc2.c_id from score sc2 where sc2.s_id = '01') and sc.s_id !='01') t on st.s_id = t.s_id

12.查询和01号学生所学课程完全相同的其他学生的学号
SQL思路：

先剔除不在01号学生选课范围内的学生
筛选选课数量与01号学生选课数量相同的结果集

SQL：
01号学生选课情况：

select c_id from score where s_id = '01'

剔除掉不在01号学生选课范围内的学生：

select sc.s_id
from score sc
where sc.c_id not in (SELECT c_id from score where s_id = '01')

剔除不在01号学生选课范围内的学号后，那么剩下的选课数量等于01号学生选课数量的就是结果集：

select sc2.s_id from score sc2 where sc2.s_id not in (select sc.s_id
from score sc
where sc.c_id not in (SELECT c_id from score where s_id = '01') ) GROUP BY sc2.s_id HAVING COUNT(sc2.s_id) = (select COUNT(c_id) from score where s_id = '01')

完整 SQL：与学生表内链接查询学生信息:

SELECT st.s_id,st.s_name FROM student st INNER JOIN (select sc2.s_id from score sc2 where sc2.s_id not in (select sc.s_id
from score sc
where sc.c_id not in (SELECT c_id from score where s_id = '01') ) GROUP BY sc2.s_id HAVING COUNT(sc2.s_id) = (select COUNT(c_id) from score where s_id = '01')) t1
ON st.s_id = t1.s_id where st.s_id !='01'

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
SQL思路：

筛选分数表中不及格的行

SELECT * FROM score sc WHERE sc.s_score < 60

筛选两门及其以上不及格课程的行

SELECT s_id FROM score sc WHERE sc.s_score < 60 GROUP BY sc.s_id HAVING COUNT(sc.s_id) >= 2

完整SQL：学生表内连接查询学生信息

SELECT st.s_id,st.s_name,avg_score FROM student st INNER JOIN (SELECT s_id,AVG(sc.s_score) as avg_score FROM score sc WHERE sc.s_score < 60 GROUP BY sc.s_id HAVING COUNT(sc.s_id) >= 2) t1 on st.s_id = t1.s_id