50道SQL练习题及答案与详细分析

发布时间：2024年01月19日

数据表介绍

--1.学生表
Student(SId,Sname,Sage,Ssex)
--SId 学生编号,Sname 学生姓名,Sage 出生年月,Ssex 学生性别

--2.课程表
Course(CId,Cname,TId)
--CId 课程编号,Cname 课程名称,TId 教师编号

--3.教师表
Teacher(TId,Tname)
--TId 教师编号,Tname 教师姓名

--4.成绩表
SC(SId,CId,score)
--SId 学生编号,CId 课程编号,score 分数

学生表 Student

create table Student(SId varchar(10),Sname varchar(10),Sage datetime,Ssex varchar(10));
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' , '1990-12-20' , '男');
insert into Student values('04' , '李云' , '1990-12-06' , '男');
insert into Student values('05' , '周梅' , '1991-12-01' , '女');
insert into Student values('06' , '吴兰' , '1992-01-01' , '女');
insert into Student values('07' , '郑竹' , '1989-01-01' , '女');
insert into Student values('09' , '张三' , '2017-12-20' , '女');
insert into Student values('10' , '李四' , '2017-12-25' , '女');
insert into Student values('11' , '李四' , '2012-06-06' , '女');
insert into Student values('12' , '赵六' , '2013-06-13' , '女');
insert into Student values('13' , '孙七' , '2014-06-01' , '女');

科目表 Course

reate table Course(CId varchar(10),Cname nvarchar(10),TId varchar(10));
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 Teacher

create table Teacher(TId varchar(10),Tname varchar(10));
insert into Teacher values('01' , '张三');
insert into Teacher values('02' , '李四');
insert into Teacher values('03' , '王五');

成绩表 SC

create table SC(SId varchar(10),CId varchar(10),score decimal(18,1));
insert into SC values('01' , '01' , 80);
insert into SC values('01' , '02' , 90);
insert into SC values('01' , '03' , 99);
insert into SC values('02' , '01' , 70);
insert into SC values('02' , '02' , 60);
insert into SC values('02' , '03' , 80);
insert into SC values('03' , '01' , 80);
insert into SC values('03' , '02' , 80);
insert into SC values('03' , '03' , 80);
insert into SC values('04' , '01' , 50);
insert into SC values('04' , '02' , 30);
insert into SC values('04' , '03' , 20);
insert into SC values('05' , '01' , 76);
insert into SC values('05' , '02' , 87);
insert into SC values('06' , '01' , 31);
insert into SC values('06' , '03' , 34);
insert into SC values('07' , '02' , 89);
insert into SC values('07' , '03' , 98);

1.查询" 01 "课程比" 02 "课程成绩高的学生的信息及课程分数

查询" 01 "课程比" 02 "课程成绩高的学生的信息及课程分数
因为需要全部的学生信息，则需要在sc表中得到符合条件的SId后与student表进行join，可以left join 也可以 right join

select * from Student RIGHT JOIN (
    select t1.SId, class1, class2 from
          (select SId, score as class1 from sc where sc.CId = '01')as t1, 
          (select SId, score as class2 from sc where sc.CId = '02')as t2
    where t1.SId = t2.SId AND t1.class1 > t2.class2
)r 
on Student.SId = r.SId;
select * from  (
    select t1.SId, class1, class2 
    from
        (SELECT SId, score as class1 FROM sc WHERE sc.CId = '01') AS t1, 
        (SELECT SId, score as class2 FROM sc WHERE sc.CId = '02') AS t2
    where t1.SId = t2.SId and t1.class1 > t2.class2
) r 
LEFT JOIN Student
ON Student.SId = r.SId;

1.1查询同时存在" 01 "课程和" 02 "课程的情况

select * from 
    (select * from sc where sc.CId = '01') as t1, 
    (select * from sc where sc.CId = '02') as t2
where t1.SId = t2.SId;

1.2查询存在" 01 "课程但可能不存在" 02 "课程的情况(不存在时显示为 null )

这一道就是明显需要使用join的情况了，02可能不存在，即为left join的右侧或right join 的左侧即可

select * from 
(select * from sc where sc.CId = '01') as t1
left join 
(select * from sc where sc.CId = '02') as t2
on t1.SId = t2.SId;
select * from 
(select * from sc where sc.CId = '02') as t2
right join 
(select * from sc where sc.CId = '01') as t1
on t1.SId = t2.SId;

1.3查询不存在" 01 "课程但存在" 02 "课程的情况

select * from sc
where sc.SId not in (
    select SId from sc 
    where sc.CId = '01'
) 
AND sc.CId= '02';

2.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

这里只用根据学生ID把成绩分组，对分组中的score求平均值，最后在选取结果中AVG大于60的即可. 注意，这里必须要给计算得到的AVG结果一个alias.（AS ss）
得到学生信息的时候既可以用join也可以用一般的联合搜索

select student.SId,sname,ss from student,(
    select SId, AVG(score) as ss from sc  
    GROUP BY SId 
    HAVING AVG(score)> 60
    )r
where student.sid = r.sid;
select Student.SId, Student.Sname, r.ss from Student right join(
      select SId, AVG(score) AS ss from sc
      GROUP BY SId
      HAVING AVG(score)> 60
)r on Student.SId = r.SId;
select s.SId,ss,Sname from(
select SId, AVG(score) as ss from sc  
GROUP BY SId 
HAVING AVG(score)> 60
)r left join 
(select Student.SId, Student.Sname from
Student)s on s.SId = r.SId;

3.查询在 SC 表存在成绩的学生信息

select DISTINCT student.*
from student,sc
where student.SId=sc.SId

4.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(没成绩的显示为 null )

select student.sid, student.sname,r.coursenumber,r.scoresum
from student,
(select sc.sid, sum(sc.score) as scoresum, count(sc.cid) as coursenumber from sc 
group by sc.sid)r
where student.sid = r.sid;

如要显示没选课的学生(显示为NULL)，需要使用join:

select s.sid, s.sname,r.coursenumber,r.scoresum
from (
    (select student.sid,student.sname 
    from student
    )s 
    left join 
    (select 
        sc.sid, sum(sc.score) as scoresum, count(sc.cid) as coursenumber
        from sc 
        group by sc.sid
    )r 
   on s.sid = r.sid
);

4.1 查有成绩的学生信息

这一题涉及到in和exists的用法，在这种小表中，两种方法的效率都差不多，但是请参考SQL查询中in和exists的区别分析
当表2的记录数量非常大的时候，选用exists比in要高效很多.
EXISTS用于检查子查询是否至少会返回一行数据，该子查询实际上并不返回任何数据，而是返回值True或False.
结论：IN()适合B表比A表数据小的情况
结论：EXISTS()适合B表比A表数据大的情况

select * from student 
where exists (select sc.sid from sc where student.sid = sc.sid);
select * from student
where student.sid in (select sc.sid from sc);

5.查询「李」姓老师的数量

select count(*)
from teacher
where tname like '李%';

6.查询学过「张三」老师授课的同学的信息

select student.* from student,teacher,course,sc
where 
    student.sid = sc.sid 
    and course.cid=sc.cid 
    and course.tid = teacher.tid 
    and tname = '张三';

7.查询没有学全所有课程的同学的信息

因为有学生什么课都没有选，反向思考，先查询选了所有课的学生，再选择这些人之外的学生.

select * from student
where student.sid not in (
  select sc.sid from sc
  group by sc.sid
  having count(sc.cid)= (select count(cid) from course)
);

8.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

这个用联合查询也可以，但是逻辑不清楚，我觉得较为清楚的逻辑是这样的：从sc表查询01同学的所有选课cid--从sc表查询所有同学的sid如果其cid在前面的结果中--从student表查询所有学生信息如果sid在前面的结果中

select * from student 
where student.sid in (
    select sc.sid from sc 
    where sc.cid in(
        select sc.cid from sc 
        where sc.sid = '01'
    )
);

9.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 * from student
where
    sid in (
    select sid from sc t1
    group by sid
    having
        group_concat(cid ORDER BY cid) = (
            select group_concat(cid ORDER BY cid) as str2 
        from sc
        where
            sid = '01')
        and sid != '01');

10.查询没学过"张三"老师讲授的任一门课程的学生姓名

仍然还是嵌套，三层嵌套，或者多表联合查询

select * from student
    where student.sid not in(
        select sc.sid from sc where sc.cid in(
            select course.cid from course where course.tid in(
                select teacher.tid from teacher where tname = "张三"
            )
        )
    );
select * from student
where student.sid not in(
    select sc.sid from sc,course,teacher 
    where
        sc.cid = course.cid
        and course.tid = teacher.tid
        and teacher.tname= "张三"
);

11.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

从SC表中选取score小于60的，并group by sid，having count 大于1
(更新采用评论1中的解法)

select student.SId, student.Sname,b.avg
from student RIGHT JOIN
(select sid, AVG(score) as avg from sc
    where sid in (
              select sid from sc 
              where score<60 
              GROUP BY sid 
              HAVING count(score)>1)
    GROUP BY sid) b on student.sid=b.sid;

12.检索" 01 "课程分数小于 60，按分数降序排列的学生信息

双表联合查询，在查询最后可以设置排序方式，语法为ORDER BY ***** DESC\ASC;

select student.*, sc.score from student, sc
where student.sid = sc.sid
and sc.score < 60
and cid = "01"
ORDER BY sc.score DESC;

13.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 *  from sc 
left join (
    select sid,avg(score) as avscore from sc 
    group by sid
    )r 
on sc.sid = r.sid
order by avscore desc;

14.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

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 
sc.CId ,
max(sc.score)as 最高分,
min(sc.score)as 最低分,
AVG(sc.score)as 平均分,
count(*)as 选修人数,
sum(case when sc.score>=60 then 1 else 0 end )/count(*)as 及格率,
sum(case when sc.score>=70 and sc.score<80 then 1 else 0 end )/count(*)as 中等率,
sum(case when sc.score>=80 and sc.score<90 then 1 else 0 end )/count(*)as 优良率,
sum(case when sc.score>=90 then 1 else 0 end )/count(*)as 优秀率 
from sc
GROUP BY sc.CId
ORDER BY count(*)DESC, sc.CId ASC

15按各科成绩进行排序，并显示排名， Score 重复时保留名次空缺

这一道题有点tricky，可以用变量，但也有更为简单的方法，即自交（左交）
用sc中的score和自己进行对比，来计算“比当前分数高的分数有几个”

select a.cid, a.sid, a.score, count(b.score)+1 as rank
from sc as a 
left join sc as b 
on a.score<b.score and a.cid = b.cid
group by a.cid, a.sid,a.score
order by a.cid, rank ASC;

16.?查询学生的总成绩，并进行排名，总分重复时不保留名次空缺

这里主要学习一下使用变量。在SQL里面变量用@来标识。

SELECT 
*,
dense_rank() over (ORDER BY sum_score DESC) AS 排名
FROM (SELECT sid,SUM(score) AS sum_score
      FROM sc
      GROUP BY SId) AS a;

17.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]，[60-0] 及所占百分比

分析题目关键词:“课程编号”、“课程名称”、“分数段”、“人数”

第一步:锁定表，成绩表、课程表

第二步:使用CASE WHEN,按照课程编号统计各分数段人数

SELECT
CId,
SUM(CASE WHEN score >= 0  AND score < 60 THEN 1 ELSE 0 END) AS '[0-60]',
SUM(CASE WHEN score >= 60 AND score < 70 THEN 1 ELSE 0 END) AS '[60-70]',
SUM(CASE WHEN score >= 70 AND score < 85 THEN 1 ELSE 0 END) AS '[70-85]',
SUM(CASE WHEN score >= 85 AND score <= 100 THEN 1 ELSE 0 END) AS '[85-100]'
FROM sc
GROUP BY CId;

第三步:转化成所占百分比

SELECT
CId,
SUM(CASE WHEN score >= 0  AND score < 60 THEN 1 ELSE 0 END) / COUNT(*) AS '[0-60]',
SUM(CASE WHEN score >= 60 AND score < 70 THEN 1 ELSE 0 END) / COUNT(*) AS '[60-70]',
SUM(CASE WHEN score >= 70 AND score < 85 THEN 1 ELSE 0 END) / COUNT(*) AS '[70-85]',
SUM(CASE WHEN score >= 85 AND score <= 100 THEN 1 ELSE 0 END) / COUNT(*) AS '[85-100]'
FROM sc
GROUP BY CId;

第四步:使用CONCAT拼接，输出百分比符号

SELECT
CId,
CONCAT(SUM(CASE WHEN score >= 0  AND score < 60 THEN 1 ELSE 0 END) / COUNT(*)*100,'%') AS '[0-60]',
CONCAT(SUM(CASE WHEN score >= 60 AND score < 70 THEN 1 ELSE 0 END) / COUNT(*)*100,'%') AS '[60-70]',
CONCAT(SUM(CASE WHEN score >= 70 AND score < 85 THEN 1 ELSE 0 END) / COUNT(*)*100,'%') AS '[70-85]',
CONCAT(SUM(CASE WHEN score >= 85 AND score <= 100 THEN 1 ELSE 0 END) / COUNT(*)*100,'%') AS '[85-100]'
FROM sc
GROUP BY CId;

第五步:关联课程表，输出课程名称

SELECT a.*,b.Cname
FROM (SELECT
      CId,
      CONCAT(SUM(CASE WHEN score >= 0  AND score < 60 THEN 1 ELSE 0 END) / COUNT(*)*100,'%') AS '[0-60]',
      CONCAT(SUM(CASE WHEN score >= 60 AND score < 70 THEN 1 ELSE 0 END) / COUNT(*)*100,'%') AS '[60-70]',
      CONCAT(SUM(CASE WHEN score >= 70 AND score < 85 THEN 1 ELSE 0 END) / COUNT(*)*100,'%') AS '[70-85]',
      CONCAT(SUM(CASE WHEN score >= 85 AND score <= 100 THEN 1 ELSE 0 END) / COUNT(*)*100,'%') AS '[85-100]'
      FROM sc
      GROUP BY CId) AS a
LEFT JOIN course AS b
ON a.CId = b.CId;

18.查询各科成绩前三名的记录

分析题目关键词，“各科成绩”、“前三名”

第一步:各科成绩排名

SELECT 
*,
dense_rank() over (PARTITION BY CId ORDER BY score DESC) AS score_rank
FROM sc;

第二步:筛选前三名

SELECT *
FROM(SELECT 
     *,
     dense_rank() over (PARTITION BY CId ORDER BY score DESC) AS score_rank
     FROM sc) AS a
WHERE score_rank <= 3;

19.查询每门课程被选修的学生数

分析题目的关键词，“每门课程”、“学生数”

第一步:锁定表，成绩表

第二步:按照课程编号进行分组统计

select cid, count(sid) from sc 
group by cid;

20.查询出只选修两门课程的学生学号和姓名

嵌套查询

select student.sid, student.sname from student
where student.sid in
(select sc.sid from sc
group by sc.sid
having count(sc.cid)=2
);

联合查询

SELECT a.SId,b.Sname
FROM(SELECT SId,COUNT(*) AS CT
     FROM sc
     GROUP BY SId
     HAVING CT = 2) AS a
LEFT JOIN student AS b
ON a.SId = b.SId;

21.查询男生、女生人数

select ssex, count(*) from student
group by ssex;

22.查询名字中含有「风」字的学生信息

select *
from student 
where student.Sname like '%风%';

23.查询同名同性学生名单，并统计同名人数

找到同名的名字并统计个数

select sname, count(*) from student
group by sname
having count(*)>1;

嵌套查询列出同名的全部学生的信息

select * from student
where sname in (
select sname from student
group by sname
having count(*)>1
);

24.查询 1990 年出生的学生名单

select *
from student
where YEAR(student.Sage)=1990;

25.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

分析题目关键词:“平均成绩”、“排序”

第一步:锁定表，成绩表

第二步:计算每门课程的平均成绩

SELECT CId,AVG(score) AS avg_score
FROM sc
GROUP BY CId;

第三步:排序(按照平均成绩降序、课程编号升序)?

SELECT CId,AVG(score) AS avg_score
FROM sc
GROUP BY CId
ORDER BY avg_score DESC,CId ASC;

26.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

分析题目关键词:“平均成绩”、“学号”、“姓名”

第一步:锁定表，成绩表、学生表

第二步:分组聚合求出每个学生平均成绩，并筛选出平均成绩大于等于85分的学生

SELECT SId,AVG(score) AS avg_score
FROM sc
GROUP BY SId
HAVING avg_score >= 85;

第三步:关联学生表，获取学生姓名?

SELECT a.sid,b.Sname,avg_score
FROM (SELECT SId,AVG(score) AS avg_score
      FROM sc
      GROUP BY SId
      HAVING avg_score >= 85) AS a
LEFT JOIN student AS b
ON a.SId = b.SId;

27.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.sname, sc.score from student, sc, course
where student.sid = sc.sid
and course.cid = sc.cid
and course.cname = "数学"
and sc.score < 60;

28.查询所有学生的课程及分数情况（存在学生没成绩，没选课的情况）

select student.sname, cid, score from student
left join sc
on student.sid = sc.sid;

29.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.sname, course.cname,sc.score from student,course,sc
where sc.score>70
and student.sid = sc.sid
and sc.cid = course.cid;

30.查询不及格的课程


SELECT DISTINCT a.Cid, a.Cname, a.Tid 
FROM course AS a 
INNER JOIN sc AS b 
ON a.Cid = b.Cid 
WHERE b.score < 60;

31.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 student.sid,student.sname 
from student,sc
where cid="01"
and score>=80
and student.sid = sc.sid;

32.求每门课程的学生人数

select sc.CId,count(*) as 学生人数
from sc
GROUP BY sc.CId;

33.成绩不重复，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

分析题目的关键词:“张三老师”、“成绩不重复”、“成绩最高”、“学生信息”、“成绩”

第一步:锁定表，教师表、课程表、成绩表、学生表

第二步:关联4张表

SELECT *
FROM sc AS a
LEFT JOIN student AS b
ON a.SId = b.SId
LEFT JOIN course AS c
ON a.CId = c.CId
LEFT JOIN teacher AS d
ON c.TId = d.TId;

第三步:增加筛选条件,[张三]老师

SELECT *
FROM sc AS a
LEFT JOIN student AS b
ON a.SId = b.SId
LEFT JOIN course AS c
ON a.CId = c.CId
LEFT JOIN teacher AS d
ON c.TId = d.TId
WHERE d.Tname = '张三';

第四步:按照分数排序，只显示第一条记录(limit)

SELECT b.*,score,Tname
FROM sc AS a
LEFT JOIN student AS b
ON a.SId = b.SId
LEFT JOIN course AS c
ON a.CId = c.CId
LEFT JOIN teacher AS d
ON c.TId = d.TId
WHERE d.Tname = '张三'
ORDER BY score DESC
LIMIT 1;

34.成绩有重复的情况下，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

分析题目的关键词:“张三老师”、“成绩重复”、“成绩最高”、“学生信息”、“成绩”

第一步:锁定表，教师表、课程表、成绩表、学生表

第二步:关联4张表，并筛选出选修[张三]老师所授课程的记录

SELECT *
FROM sc AS a
LEFT JOIN student AS b
ON a.SId = b.SId
LEFT JOIN course AS c
ON a.CId = c.CId
LEFT JOIN teacher AS d
ON c.TId = d.TId
WHERE d.Tname = '张三';

第三步:使用dense_rank()进行排名

SELECT 
b.*，a.score,d.Tname,
dense_rank() over (ORDER BY score DESC) AS score_rank
FROM sc AS a
LEFT JOIN student AS b
ON a.SId = b.SId
LEFT JOIN course AS c
ON a.CId = c.CId
LEFT JOIN teacher AS d
ON c.TId = d.TId
WHERE d.Tname = '张三';

第四步:筛选出第1名

SELECT *
FROM(SELECT 
     b.*，a.score,d.Tname,
     dense_rank() over (ORDER BY score DESC) AS score_rank
     FROM sc AS a
     LEFT JOIN student AS b
     ON a.SId = b.SId
     LEFT JOIN course AS c
     ON a.CId = c.CId
     LEFT JOIN teacher AS d
     ON c.TId = d.TId
     WHERE d.Tname = '张三') 
WHERE score_rank = 1;

35.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

分析题目的关键词:“不同课程”、“成绩相同”

第一步:锁定表，成绩表

第二步:自连接成绩表，连接条件是学生编号相同、课程编号不同、成绩相同

SELECT a.*
FROM sc AS a
INNER JOIN sc AS b
ON a.SId = b.SId AND a.CId != b.CId AND a.score = b.score;

第三步:去重

SELECT DISTINCT a.*
FROM sc AS a
INNER JOIN sc AS b
ON a.SId = b.SId AND a.CId != b.CId AND a.score = b.score;

36.查询每门功成绩最好的前两名

分析题目的关键词:“每门科目”、“成绩最好的前两名”

第一步:锁定表，成绩表

第二步:使用dense_rank()排名

SELECT
*,
dense_rank() over (PARTITION BY CId ORDER BY score DESC) AS score_rank
FROM sc;

第三步:筛选出前两名

SELECT *
FROM (SELECT
      *,
      dense_rank() over (PARTITION BY CId ORDER BY score DESC) AS score_rank
      FROM sc) AS a
WHERE score_rank <= 2;

37.统计每门课程的学生选修人数（超过 5 人的课程才统计）。

select sc.cid, count(sid) as cc from sc
group by cid
having cc >5;

38.检索至少选修两门课程的学生学号

select sid, count(cid) as cc from sc
group by sid
having cc>=2;

39.查询选修了全部课程的学生信息

分析题目的关键词:“选修全部课程”、“学生信息”

第一步:锁定表，学生表、成绩表、课程表

第二步:求出课程表的总课程数

SELECT COUNT(*) FROM course;

第三步:求出每名学生的选修课程数

SELECT SId,COUNT(*)
FROM sc
GROUP BY SId;

第四步:筛选出学习全部课程的学生学生SId

SELECT SId,COUNT(*)
FROM sc
GROUP BY SId
HAVING COUNT(*) = (SELECT COUNT(*) FROM course);

第五步:使用子查询，获取学生信息

SELECT *
FROM student
WHERE SId IN(SELECT SId
             FROM sc
             GROUP BY SId
             HAVING COUNT(*) = (SELECT COUNT(*) FROM course));

40.查询各学生的年龄，只按年份来算

SELECT Sid,Sname,YEAR(NOW())-YEAR(sage) 年龄 FROM Student

41.按照出生日期来算，当前月日 < 出生年月的月日则，年龄减一

分析题目的关键词:“年龄”、“当前月日<出生年月的月日，则年龄减一”

第一步:锁定表，学生表

第二步:使用NOW()、TIMESTAMPDIFF()函数计算年龄

SELECT *,TIMESTAMPDIFF(YEAR,Sage,NOW()) AS age
FROM student;

42.查询本周过生日的学生

select *
from student 
where WEEKOFYEAR(student.Sage)=WEEKOFYEAR(CURDATE());

43.查询下周过生日的学生

select *
from student 
where WEEKOFYEAR(student.Sage)=WEEKOFYEAR(CURDATE())+1;

44.查询本月过生日的学生

select *
from student 
where MONTH(student.Sage)=MONTH(CURDATE());

45.查询下月过生日的学生

select *
from student 
where MONTH(student.Sage)=MONTH(CURDATE())+1;

文章来源:https://blog.csdn.net/Fithck/article/details/135684570
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50道SQL练习题及答案与详细分析

学生表 Student

科目表 Course

教师表 Teacher

成绩表 SC

1.查询" 01 "课程比" 02 "课程成绩高的学生的信息及课程分数

1.1查询同时存在" 01 "课程和" 02 "课程的情况

1.2查询存在" 01 "课程但可能不存在" 02 "课程的情况(不存在时显示为 null )

1.3查询不存在" 01 "课程但存在" 02 "课程的情况

2.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

3.查询在 SC 表存在成绩的学生信息

4.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(没成绩的显示为 null )

4.1 查有成绩的学生信息

5.查询「李」姓老师的数量

6.查询学过「张三」老师授课的同学的信息

7.查询没有学全所有课程的同学的信息

8.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

9.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

10.查询没学过"张三"老师讲授的任一门课程的学生姓名

11.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

12.检索" 01 "课程分数小于 60，按分数降序排列的学生信息

13.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

14.查询各科成绩最高分、最低分和平均分：

15按各科成绩进行排序，并显示排名， Score 重复时保留名次空缺

16.?查询学生的总成绩，并进行排名，总分重复时不保留名次空缺

17.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]，[60-0] 及所占百分比

18.查询各科成绩前三名的记录

19.查询每门课程被选修的学生数

20.查询出只选修两门课程的学生学号和姓名

21.查询男生、女生人数

22.查询名字中含有「风」字的学生信息

23.查询同名同性学生名单，并统计同名人数

24.查询 1990 年出生的学生名单

25.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

26.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

27.查询课程名称为「数学」，且分数低于 60 的学生姓名和分数

28.查询所有学生的课程及分数情况（存在学生没成绩，没选课的情况）

29.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

30.查询不及格的课程

31.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

32.求每门课程的学生人数

33.成绩不重复，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

34.成绩有重复的情况下，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

35.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

36.查询每门功成绩最好的前两名

37.统计每门课程的学生选修人数（超过 5 人的课程才统计）。

38.检索至少选修两门课程的学生学号

39.查询选修了全部课程的学生信息

40.查询各学生的年龄，只按年份来算

41.按照出生日期来算，当前月日 < 出生年月的月日则，年龄减一

42.查询本周过生日的学生

43.查询下周过生日的学生

44.查询本月过生日的学生

45.查询下月过生日的学生

9.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