SQL面试必会50道题实操

Fun.ny

已于 2025-01-03 11:32:12 修改

阅读量263

点赞数 1

CC 4.0 BY-SA版权

文章标签： sql 数据库 java

于 2024-11-21 16:43:56 首次发布

本文链接：https://blog.youkuaiyun.com/nianiaooa/article/details/143677775

sql实操案例

一、导入数据

--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);

二、50道面试题原题（先自己实操一遍）

1、查询课程编号为“01”的课程比“02”的课程成绩高的所有学生的学号（难）
2、查询平均成绩大于60分的学生的学号和平均成绩
3、查询所有学生的学号、姓名、选课数、总成绩
4、查询姓“猴”的老师的个数
5、查询没学过“张三”老师课的学生的学号、姓名
6、查询学过“张三”老师所教的所有课的同学的学号、姓名
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
8、查询课程编号为“02”的总成绩
9、查询所有课程成绩小于60分的学生的学号、姓名
10、查询没有学全所有课的学生的学号、姓名
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的学号和姓名 （难）
12、查询和“01”号同学所学课程完全相同的其他同学的学号（难）
13、查询没学过"张三"老师讲授的任一门课程的学生姓名 和47题一样
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
16、检索"01"课程分数小于60，按分数降序排列的学生信息
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(难)
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
19、按各科成绩进行排序，并显示排名(难)
20、查询学生的总成绩并进行排名
21、查询不同老师所教不同课程平均分从高到低显示
22、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（重要 25类似）
23、使用分段[100-85],[85-70],[70-60],[<60]来统计各科成绩，分别统计各分数段人数：课程ID和课程名称
24、查询学生平均成绩及其名次
25、查询各科成绩前三名的记录（不考虑成绩并列情况）
26、查询每门课程被选修的学生数
27、查询出只有两门课程的全部学生的学号和姓名
28、查询男生、女生人数
29、查询名字中含有"风"字的学生信息
31、查询1990年出生的学生名单
32、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
33、查询每门课程的平均成绩，结果按平均成绩升序排序，平均成绩相同时，按课程号降序排列
34、查询课程名称为"数学"，且分数低于60的学生姓名和分数
35、查询所有学生的课程及分数情况
36、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
37、查询不及格的课程并按课程号从大到小排列
38、查询课程编号为03且课程成绩在80分以上的学生的学号和姓名
39、求每门课程的学生人数
40、查询选修“张三”老师所授课程的学生中成绩最高的学生姓名及其成绩
41、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 （难）
42、查询每门功成绩最好的前两名
43、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
44、检索至少选修两门课程的学生学号
45、查询选修了全部课程的学生信息
46、查询各学生的年龄
47、查询没学过“张三”老师讲授的任一门课程的学生姓名
48、查询两门以上不及格课程的同学的学号及其平均成绩
49、查询本月过生日的学生
50、查询下一个月过生日的学生

三、补充知识点

case when用法

CASE WHEN score IS NULL THEN '缺席考试' ELSE '正常' END

# 有分数score，score<60返回不及格，score>=60返回及格，score>=80返回优秀
SELECT
    stu_name,
    (CASE WHEN stu_score < 60 THEN '不及格'
        WHEN stu_score >= 60 AND stu_score < 80 THEN '及格'
        WHEN stu_score >= 80 THEN '优秀'
        ELSE '异常' END) AS REMARK
FROM students;

rank()用法

select s_id,s_score,rank() over(order by s_score desc ) as ranks
from score where c_id='01';

四、代码实操

##简单单表查询
#4、查询姓“李”的老师的个数
select count(t_id) from teacher where t_name like '李%';

#29、查询名字中含有"风"字的学生信息
select * from student
where s_name like '%风%';

#26、查询每门课程被选修的学生数
select c_id,count(s_id) from score
group by c_id;

#28、查询男生、女生人数
#练习写法
select s_sex,count(s_sex) from student
group by s_sex;
#参考写法
select (select count(*) from student where s_sex = '男') as "男生数",
(select count(*) from student where s_sex = '女') as "女生数"

#31、查询1990年出生的学生名单
#练习写法
select s_name from student
where s_birth like '1990%';
#参考写法
select * from student
where year(s_birth) = 1990 ;

#39、求每门课程的学生人数
select c_id,count(s_id) from score group by c_id;

#46、查询各学生的年龄
#练习写法
select s_name,s_birth,
floor(datediff(current_date,s_birth)/365) as age 
from student;
#参考写法1：year() 直接取得年份
select st.*, 2024 - year(st.s_birth) as "年龄"
from student as st
#参考写法二：round()四舍五入,now()现在的时间,datadiff(日期1，日期2),返回日期2-日期1的天数
select st.*, round(datediff(now(),st.s_birth)/365) as "年龄"
from student as st
#参考写法三：floor()向下取整
select st.*, floor(datediff(now(),st.s_birth)/365) as "年龄"
from student as st

#49、查询本月过生日的学生
select s_name,s_birth from student
where month(s_birth)=month(current_date);

#50、查询下一个月过生日的学生
select s_name,s_birth from student
where month(s_birth)=month(current_date)+1;

#8、查询课程编号为“02”的总成绩
#练习写法
select sum(s_score) from score where c_id='02';
#参考写法
select c_id, sum(s_score)
from score
group by c_id
having c_id = '02'

#查询课程编号为“02”的平均成绩
select avg(s_score) from score where c_id='02';

#20、查询学生的总成绩并进行排名
#练习写法
select sum(s_score) from score 
group by s_id
order by sum(s_score);
#参考写法
-- 考察点：窗口函数 rank()
select a.s_id,a.s_name,
(case when sum(s_score) is Null then 0 else sum(s_score) end) as "总分",
rank() over (order by sum(s_score) desc) as "rank"
from student as a left join score as b
on a.s_id = b.s_id
group by a.s_id

#24、查询学生平均成绩及其名次
#练习写法（需重写）
select avg(s_score) 
from score 
group by s_id
order by avg(s_score) desc;#名次部分还需要显示行数；
#参考写法
-- 考察点：row_number 的用法,记得里面可以指定是升序还是降序排列
select a.s_id,avg(s_score),row_number() over (order by avg(s_score) desc) as "rank"
from student as a,score as b
where a.s_id = b.s_id
group by a.s_id
order by avg(s_score) desc
#row_number的用法需学习；

#2、查询
平均成绩大于60分的学生的学号和平均成绩
select s_id,avg(s_score) as s_avg
from score
group by s_id
having s_avg>60;#分组后筛选的结果用having聚合函数
#参考写法
-- 考察点：聚集函数avg()，数据分组groupby，过滤分组
select s_id,avg(s_score) as "平均成绩" 
from score
group by s_id
having 平均成绩 > 60

#44、检索至少选修两门课程的学生学号
select s_id,count(c_id) from score 
group by s_id 
having count(c_id)>=2;
#参考写法
select s_id from score
group by s_id
having count(*) >= 2

#33、查询每门课程的平均成绩，结果按平均成绩升序排序，平均成绩相同时，按课程号降序排列
select avg(s_score) from score
group by c_id
order by avg(s_score) asc,c_id desc;
#参考写法
select c_id,avg(s_score) as avg_score
from score
group by c_id
order by avg_score asc,c_id desc

##要调用其它表格条件的单表查询
#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
select s_id,s_name from student where
s_id in(select s_id from score where c_id='01'
intersect #取交集
select s_id from score where c_id='02');
#参考写法
-- 思路：和第1题类似，将子表连接起来
select distinct s.s_id,st.s_name 
from score as s
inner join (select * from score where c_id = '01') as s1 on s.s_id = s1.s_id
inner join (select * from score where c_id = '02') as s2 on s.s_id = s2.s_id
inner join student as st on s.s_id = st.s_id
##distinct的意思待了解

#9、查询所有课程成绩小于60分的学生的学号、姓名
select s_id,s_name from student
where s_id not in(select s_id from score where s_score>60);#如果一门课都没有选的怎么办？
#参考写法
-- 思路：最大的一门课都小于60，也就是所有的都小于60了，另外，有的同学一门课都没有考过试，所以我后面多加了一个姓名不在成绩表里的选项
select a.s_id,a.s_name 
from student as a 
where (select max(s_score) from score group by s_id having s_id = a.s_id ) < 60 
or a.s_id not in (select distinct s_id from score)

#10、查询没有学全所有课的学生的学号、姓名
select s_id,s_name from student where
s_id not in(select s_id from score where c_id='01'
intersect
select s_id from score where c_id='02'
intersect
select s_id from score where c_id='03');
#参考写法
-- 考察点：过滤分组,王菊一门课都没有选，但是也应该是答案，所以要用左连接
select a.s_id,a.s_name
from student as a left join score as b
on a.s_id = b.s_id
group by a.s_id
having count(b.c_id) < (select count(distinct c_id) from score)


#5、查询没学过“张三”老师课的学生的学号、姓名
select c_id from course where t_id in(select t_id from teacher where t_name ='张三' );#先查询出来张三教的课程号为'02';
select s_id,s_name from student where
s_id not in(select s_id from score where c_id='02');
#参考写法
-- 考察点：not in
-- 思路：先找出学过“张三”老师课的学生学号，然后利用反向排除法
select s_id,s_name 
from student where s_id not in (
select s.s_id
from student as s
inner join score as sc on s.s_id = sc.s_id
inner join course as c on c.c_id = sc.c_id
inner join teacher as t on t.t_id = c.t_id
where t_name = '张三')

#6、查询学过“张三”老师所教的所有课的同学的学号、姓名
select c_id from course where t_id in(select t_id from teacher where t_name ='张三' );#先查询出来张三教的课程号为'02';
select s_id,s_name from student where
s_id in(select s_id from score where c_id='02');

#47、查询没学过“张三”老师讲授的任一门课程的学生姓名
#13、查询没学过"张三"老师讲授的任一门课程的学生姓名 和47题一样

#45、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(c_id)>=3);

#27、查询出只有两门课程的全部学生的学号和姓名
select s_id,s_name from student where s_id
in(select s_id from score group by s_id having count(c_id)=2);

#38、查询课程编号为03且课程成绩在80分以上的学生的学号和姓名
select s_id,s_name from student where
s_id in(select s_id from score where c_id='03' and s_score>80);

#48、查询两门以上不及格课程的同学的学号及其平均成绩
select s_id,avg(s_score) from score where s_score<60
group by s_id
having count(c_id)>=2;

#43、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(c_id) from score
group by c_id 
having count(c_id)>5
order by count(c_id) desc,c_id asc;

#37、查询不及格的课程并按课程号从大到小排列
select c_id,s_score from score
where s_score<60
order by c_id asc;

##简单联合查询
#3、查询所有学生的学号、姓名、选课数、总成绩
select student.s_id,student.s_name,
count(score.c_id) as total_courses,
sum(score.s_score) as total_score
from student
left join score
on student.s_id = score.s_id
group by student.s_id;
#32、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 student.s_id,student.s_name,avg(score.s_score) from student
left join score on student.s_id = score.s_id
group by student.s_id
having avg(score.s_score)>=85; 

#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 student.s_id,student.s_name,avg(score.s_score)
from score
left join student
on student.s_id=score.s_id
where score.s_score<60 #联合查询时条件放join on后面
group by score.s_id
having count(score.c_id)>=2;

#34、查询课程名称为"数学"，且分数低于60的学生姓名和分数
select student.s_name,score.s_score from student
inner join
score on score.s_id=student.s_id
inner join
course on score.c_id=course.c_id
where course.c_name='数学' and score.s_score<60;

#16、检索"01"课程分数小于60，按分数降序排列的学生信息
select student.*,score.s_score from student
inner join score
on score.s_id=student.s_id
where score.c_id='01' and score.s_score <60
order by score.s_score desc;

#40、查询选修“张三”老师所授课程的学生中成绩最高的学生姓名及其成绩
select student.s_name,score.s_score from student
inner join score
on student.s_id=score.s_id
inner join course
on score.c_id=course.c_id
inner join teacher
on course.t_id = teacher.t_id
where teacher.t_name='张三'
order by score.s_score desc limit 1;

#21、查询不同老师所教不同课程平均分从高到低显示
select teacher.t_name,avg(score.s_score) as avg_score from score
inner join course on
course.c_id=score.c_id
inner join teacher on 
teacher.t_id = course.t_id
group by score.c_id
order by avg(score.s_score) desc;

#25、查询各科成绩前三名的记录（不考虑成绩并列情况）
select course.c_name,score.s_score from score
inner join course
on course.c_id = score.c_id
where course.c_name ='语文'
order by score.s_score desc limit 3;

select course.c_name,score.s_score from score
inner join course
on course.c_id = score.c_id
where course.c_name ='数学'
order by score.s_score desc limit 3;

select course.c_name,score.s_score from score
inner join course
on course.c_id = score.c_id
where course.c_name ='英语'
order by score.s_score desc limit 3;

#42、查询每门功成绩最好的前两名
select course.c_name,score.s_score from score
inner join course
on course.c_id = score.c_id
where course.c_name ='语文'
order by score.s_score desc limit 2;

select course.c_name,score.s_score from score
inner join course
on course.c_id = score.c_id
where course.c_name ='数学'
order by score.s_score desc limit 2;

select course.c_name,score.s_score from score
inner join course
on course.c_id = score.c_id
where course.c_name ='英语'
order by score.s_score desc limit 2;


#35、查询所有学生的课程及分数情况
select student.s_name,course.c_name,score.s_score from student
inner join score on
score.s_id = student.s_id
inner join course on
course.c_id = score.c_id;

#36、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 student.s_name,course.c_name,score.s_score from student
inner join score on
score.s_id = student.s_id
inner join course on
course.c_id = score.c_id
where score.s_score>70;


#22、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（重要 25类似）
select course.c_name,score.s_score from score
inner join course
on course.c_id = score.c_id
where course.c_name ='语文'
order by score.s_score desc limit 1,2;

select course.c_name,score.s_score from score
inner join course
on course.c_id = score.c_id
where course.c_name ='数学'
order by score.s_score desc limit 1,2;

select course.c_name,score.s_score from score
inner join course
on course.c_id = score.c_id
where course.c_name ='英语'
order by score.s_score desc limit 1,2;


#23、使用分段[100-85],[85-70],[70-60],[<60]来统计各科成绩，分别统计各分数段人数：课程ID和课程名称
select course.c_id,course.c_name,
sum(case when score.s_score between 85 and 100 then 1 else 0 end ) as '85-100',
sum(case when score.s_score between 70 and 85 then 1 else 0 end ) as '70-85',
sum(case when score.s_score between 60 and 70 then 1 else 0 end ) as '60-70',
sum(case when score.s_score <60 then 1 else 0 end ) as '<60'
from score
inner join course on
course.c_id = score.c_id
group by score.c_id;

#1、查询课程编号为“01”的课程比“02”的课程成绩高的所有学生的学号（难）
SELECT a.s_id
FROM (SELECT s_id, c_id, s_score FROM score WHERE c_id='01') a
INNER JOIN (SELECT s_id, c_id, s_score FROM score WHERE c_id='02') b
ON a.s_id = b.s_id
WHERE a.s_score > b.s_score;

#12、查询和“01”号同学所学课程完全相同的其他同学的学号（难）
SELECT s_id FROM Score WHERE c_id 
IN (SELECT c_id FROM Score WHERE s_id = '01' )
AND s_id != '01'
GROUP BY s_id
HAVING COUNT(c_id) = (SELECT COUNT(c_id) FROM Score WHERE s_id = '01');

#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的学号和姓名 （难）
SELECT s_id,s_name FROM Student WHERE s_id IN (
    SELECT s_id FROM Score WHERE c_id IN (
        SELECT c_id FROM Score WHERE s_id = '01'
    ) AND s_id != '01'
    GROUP BY s_id
);


#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(难)
#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
#19、按各科成绩进行排序，并显示排名(难)
#41、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 （难）