数据库面试题（4）常见SQL编程

2023年4月30日倾城 Comments 0条评论 24 人阅读

表设计

（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

--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);

编程题

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

或者：

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<c.s_score

3、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 b.s_id,b.s_name,ROUND(AVG(a.s_score),2) as avg_score from 
  student b 
  join score a on b.s_id = a.s_id
  GROUP BY b.s_id,b.s_name HAVING avg_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')

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

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)

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

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')

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  s_id having count(1)>=2)
  GROUP BY a.s_id,a.s_name

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;

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 then 1 else 0 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 then 1 else 0 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

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

(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,
t1.s_score,
(select count(distinct t2.s_score) from score t2 where t2.s_score>=t1.s_score and t2.c_id='02') rank
FROM score t1 where t1.c_id='02'
order by t1.s_score desc) t2)
union
(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='03') rank
FROM score t1 where t1.c_id='03'
order by t1.s_score desc) t3)

20、查询学生的总成绩并进行排名

select a.s_id,
  @i:=@i+1 as i,
  @k:=(case when @score=a.sum_score then @k else @i end) as rank,
  @score:=a.sum_score as score
from (select s_id,SUM(s_score) as sum_score from score GROUP BY s_id ORDER BY sum_score DESC)a,
  (select @k:=0,@i:=0,@score:=0)s

21、查询不同老师所教不同课程平均分从高到低显示

  select a.t_id,c.t_name,a.c_id,ROUND(avg(s_score),2) as avg_score from course a
left join score b on a.c_id=b.c_id 
left join teacher c on a.t_id=c.t_id
GROUP BY a.c_id,a.t_id,c.t_name ORDER BY avg_score DESC;

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

select d.*,c.排名,c.s_score,c.c_id from (
select a.s_id,a.s_score,a.c_id,@i:=@i+1 as 排名 from score a,(select @i:=0)s where a.c_id='01'  
ORDER BY a.s_score DESC  
)c
left join student d on c.s_id=d.s_id
where 排名 BETWEEN 2 AND 3
UNION
select d.*,c.排名,c.s_score,c.c_id from (
select a.s_id,a.s_score,a.c_id,@j:=@j+1 as 排名 from score a,(select @j:=0)s where a.c_id='02'  
ORDER BY a.s_score DESC
)c
left join student d on c.s_id=d.s_id
where 排名 BETWEEN 2 AND 3
UNION
select d.*,c.排名,c.s_score,c.c_id from (
select a.s_id,a.s_score,a.c_id,@k:=@k+1 as 排名 from score a,(select @k:=0)s where a.c_id='03' 
ORDER BY a.s_score DESC
)c
left join student d on c.s_id=d.s_id
where 排名 BETWEEN 2 AND 3;

23、统计各科成绩各分数段人数、课程编号、课程名称、[100-85],[85-70],[70-60],[0-60]及所占百分比

select distinct f.c_name,a.c_id,b.`85-100`,b.百分比,c.`70-85`,c.百分比,d.`60-70`,d.百分比,e.`0-60`,e.百分比 from score a
left join (select c_id,SUM(case when s_score >85 and s_score <=100 then 1 else 0 end) as `85-100`,
  ROUND(100*(SUM(case when s_score >85 and s_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 s_score >70 and s_score <=85 then 1 else 0 end) as `70-85`,
  ROUND(100*(SUM(case when s_score >70 and s_score <=85 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 s_score >60 and s_score <=70 then 1 else 0 end) as `60-70`,
  ROUND(100*(SUM(case when s_score >60 and s_score <=70 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 (select c_id,SUM(case when s_score >=0 and s_score <=60 then 1 else 0 end) as `0-60`,
  ROUND(100*(SUM(case when s_score >=0 and s_score <=60 then 1 else 0 end)/count(*)),2) as 百分比
from score GROUP BY c_id)e on a.c_id=e.c_id
left join course f on a.c_id = f.c_id

24、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(s_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;

25、查询各科成绩前三名的记录

--选出b表比a表成绩大的所有组
--选出比当前id成绩大的小于三个的
select a.s_id,a.c_id,a.s_score from score a 
  left join score b on a.c_id = b.c_id and a.s_score<b.s_score
  group by a.s_id,a.c_id,a.s_score HAVING COUNT(b.s_id)<3
  ORDER BY a.c_id,a.s_score DESC

26、查询每门课程被选修的学生数

select c_id,count(s_id) from score a GROUP BY c_id

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);

28、查询男生、女生人数

select s_sex,COUNT(s_sex) as 人数 from student GROUP BY s_sex

29、查询名字中含有"风"字的学生信息

select * from student where s_name like '%风%';

30、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

31、查询1990年出生的学生名单

select s_name from student where s_birth like '1990%'

32、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(s_score),2) as avg_score from score GROUP BY c_id ORDER BY avg_score DESC,c_id ASC

33、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.s_score),2) as avg_score from score a
left join student b on a.s_id=b.s_id GROUP BY s_id HAVING avg_score>=85

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 a.s_name,b.s_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 ='数学') and b.s_score<60

35、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 '语文' then b.s_score else 0 end) as '语文',
  SUM(case c.c_name when '数学' then b.s_score else 0 end) as '数学',
  SUM(case c.c_name when '英语' then b.s_score else 0 end) as '英语',
  SUM(b.s_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

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 a.s_name,b.c_name,c.s_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.s_score>=70

37、查询不及格的课程

select a.s_id,a.c_id,b.c_name,a.s_score from score a left join course b on a.c_id = b.c_id
  where a.s_score<60

38、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.s_score>80

39、求每门课程的学生人数

select count(*) from score GROUP BY c_id;

40、查询选修"张三"老师所授课程的学生中，成绩最高的学生信息及其成绩

--查询老师id  
select c_id from course c,teacher d where c.t_id=d.t_id and d.t_name='张三'
--查询最高分（可能有相同分数）
select MAX(s_score) from score where c_id='02'
--最终查询语句
select a.*,b.s_score,b.c_id,c.c_name 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
  where b.c_id =(select c_id from course c,teacher d where c.t_id=d.t_id and d.t_name='张三')
  and b.s_score in (select MAX(s_score) from score where c_id='02')

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

select DISTINCT b.s_id,b.c_id,b.s_score from score a,score b where a.c_id != b.c_id and a.s_score = b.s_score
  ```

**42、查询每门功成绩最好的前两名**

```sql
select a.s_id,a.c_id,a.s_score from score a
where (select COUNT(1) from score b where b.c_id=a.c_id and b.s_score>=a.s_score)<=2 ORDER BY a.c_id

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(*) as total from score GROUP BY c_id HAVING total>5 ORDER BY total,c_id ASC

44、检索至少选修两门课程的学生学号

select s_id,count(*) as sel from score GROUP BY s_id HAVING sel>=2

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(*)=(select count(*) from course))

46、查询各学生的年龄

--按照出生日期来算，当前月日 < 出生年月的月日则，年龄减一
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;

47、查询本周过生日的学生

select * from student where WEEK(DATE_FORMAT(NOW(),'%Y%m%d'))=WEEK(s_birth)
select * from student where YEARWEEK(s_birth)=YEARWEEK(DATE_FORMAT(NOW(),'%Y%m%d'))

48、查询下周过生日的学生

select * from student where WEEK(DATE_FORMAT(NOW(),'%Y%m%d'))+1 =WEEK(s_birth)

49、查询本月过生日的学生

select * from student where MONTH(DATE_FORMAT(NOW(),'%Y%m%d')) =MONTH(s_birth)

50、查询下月过生日的学生

select * from student where MONTH(DATE_FORMAT(NOW(),'%Y%m%d'))+1 =MONTH(s_birth)

参考
https://mp.weixin.qq.com/s?__biz=MzIyMDQ4Mjg3Mw==&mid=2247494858&idx=1&sn=8d2b4f53ebdc4579e061722fb8638102&chksm=97c9f09ca0be798a3f1fb3ddc0e506e9587a0c10bce631b3cf1787a1c9da5bb436a28caf36ed&scene=27

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数据库面试题（4）常见SQL编程

2023年4月30日倾城 Comments 0条评论 24 人阅读

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