MySQL面试题

面试题一

 

 1、创建一个数据库

create database db_one;

2、 创建四张表

create table student(

        s_id int(10) not null comment '学号' primary key,

        s_name varchar(20) not null comment '姓名',

        s_birth  year comment '生日',

        s_sex varchar(4) default "女" comment '性别'

)engine = innodb;

create table teacher(

        t_id int(10) not null comment '教师编号' primary key,

        t_name varchar(20) not null comment '姓名'

)engine = innodb;

create table course(

        c_id int(10) not null comment '课程编号' primary key,

        c_name varchar(20) not null comment '课程名称',

        t_id int(10) comment '教师编号',

        foreign key(t_id) references teacher(t_id) 

)engine = innodb;

create table score(

        s_id int(10) comment '学号',

        c_id int(10) comment '课程编码',

        s_score int(10) comment '成绩',

        foreign key(s_id) references student(s_id),

        foreign key(c_id) references course(c_id) 

)engine = innodb;

3、插入数据

向student表插入记录的INSERT语句如下：

INSERT INTO student VALUES( 901,'李大',2000, '男');
INSERT INTO student VALUES( 902,'李二',2001, '男');
INSERT INTO student VALUES( 903,'李三',2002, '女');
INSERT INTO student VALUES( 904,'李四',2000, '男');
INSERT INTO student VALUES( 905,'王五',2001, '女');
INSERT INTO student VALUES( 906,'王六',2003, '男');

向teacher表插入记录的INSERT语句如下：

INSERT INTO teacher VALUES( 101,'张一');
INSERT INTO teacher VALUES( 102,'张二');
INSERT INTO teacher VALUES( 103,'张三');
INSERT INTO teacher VALUES( 104,'张四');

向course表插入记录的INSERT语句如下：

INSERT INTO course VALUES( 1001,'数学',101);
INSERT INTO course VALUES( 1002,'英语',102);
INSERT INTO course VALUES( 1003,'中文',103);
INSERT INTO course VALUES( 1004,'计算机',104);

向score表插入记录的INSERT语句如下：
INSERT INTO score VALUES(901, 1004,98);
INSERT INTO score VALUES(901, 1002,80);
INSERT INTO score VALUES(902, 1004,65);
INSERT INTO score VALUES(902, 1003,88);
INSERT INTO score VALUES(903, 1003,95);

INSERT INTO score VALUES(903, 1004,95);
INSERT INTO score VALUES(904, 1004,70);
INSERT INTO score VALUES(904, 1002,92);
INSERT INTO score VALUES(905, 1002,94);
INSERT INTO score VALUES(906, 1002,90);
INSERT INTO score VALUES(906, 1004,85);

4、查询

1、 select student.s_id,student.s_name,count( score.c_id ) as '选课数',sum( score.s_score ) as '总成绩'  from student inner join score on student.s_id = score.s_id group by score.s_id;

2、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'));

select s_id,s_name from student where s_id 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')));

3、select c_id from score where s_id='902'；

select sno from sc group by sno having count(*)=(select count(*)nfrom sc where sno='1002');

select sc1.s_id from (select s_id from score group by s_id having count(*)=(select count(*) n  from score where s_id='902')) sc1,score where sc1.s_id=score.s_id and c_id in (select c_id from score where s_id='902') group by score.s_id having count(*)=(select count(*) from score where s_id='902');

4、select s_id,avg(s_score) as avgscore from score group by s_id;

select st.s_id,st.s_name,co.c_name,avgscore from score sc,course co,student st,(select s_id,avg(s_score) as avgscore from score group by s_id) as nb where sc.c_id=co.c_id and sc.s_id=st.s_id and nb.s_id=st.s_id order by avgscore desc;

5、select a.c_id,b.c_name,sum(case when s_score between 85 and 100 then 1 else 0 end ) as '[100-85]',sum(case when s_score >= 70 and s_score <85 then 1 else 0 end ) as '[85-70]',sum(case when s_score >= 60 and s_score <70 then 1 else 0 end ) as '[70-60]',sum(case when s_score <60 then 1 else 0 end ) as '[<60]' from score as a right join course as b on a.c_id=b.c_id group by a.c_id,b.c_name;

面试题二

 

1、创建数据库

 create database db_two;

2、创建三个表

create table departments(

        dept_no int(10) not null primary key,

        dept_name varchar(20) not null 

)engine = innodb;

create table dept_emp(

        emp_no int(10) not null  primary key,

        dept_no int(10) not null,

        from_date date,

        to_date date,

        foreign key(dept_no) references departments(dept_no)

)engine = innodb;

create table salaries(

        salary int(10) not null primary key,

        emp_no int(10) not null,

        from_date date,

        to_date date,

        foreign key(emp_no) references dept_emp(emp_no)

)engine = innodb;

3、 添加数据

向departments表插入记录的INSERT语句如下：

INSERT INTO departments VALUES( 001,'Marketing');
INSERT INTO departments VALUES( 002,'Finance');

向dept_emp表插入记录的INSERT语句如下：

INSERT INTO dept_emp VALUES( 10001,001,'2001-06-22','9999-01-01');
INSERT INTO dept_emp VALUES( 10002,001,'1996-08-03','9999-01-01');
INSERT INTO dept_emp VALUES( 10003,002,'1996-08-03','9999-01-01');

向salaries表插入记录的INSERT语句如下：

INSERT INTO salaries VALUES( 85097,10001,'2001-06-22','2002-06-22');

INSERT INTO salaries VALUES( 88958,10001,'2002-06-22','9999-01-01');
INSERT INTO salaries VALUES( 72527,10002,'1996-08-03','9999-01-01');
INSERT INTO salaries VALUES( 32323,10003,'1996-08-03','9999-01-01');

4、查看输出

 select dept_no,dept_name,sum from departments d,salsries s group by dept_no having count()；

50个题

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

create 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);

练习题

1.查询" 01 "课程比" 02 "课程成绩高的学生的信息及课程分数

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';

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;

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、所有课程的成绩总和
联合查询不会显示没选课的学生：

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

);

4.2 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);

查询「李」姓老师的数量
select count(*)

from teacher

where tname like '李%';

查询学过「张三」老师授课的同学的信息
多表联合查询
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';

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)

);

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
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
不会做。

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;

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;

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;

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

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

group by a.cid, a.sid,a.score

order by a.cid, rank ASC;

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。
set @crank=0;

select q.sid, total, @crank := @crank +1 as rank from(

select sc.sid, sum(sc.score) as total from sc

group by sc.sid

order by total desc)q;

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
有时候觉得自己真是死脑筋。group by以后的查询结果无法使用别名，所以不要想着先单表group by计算出结果再从第二张表里添上课程信息，而应该先将两张表join在一起得到所有想要的属性再对这张总表进行统计计算。这里就不算百分比了，道理相同。
注意一下，用case when 返回1 以后的统计不是用count而是sum
select course.cname, course.cid,

sum(case when sc.score<=100 and sc.score>85 then 1 else 0 end) as "[100-85]",

sum(case when sc.score<=85 and sc.score>70 then 1 else 0 end) as "[85-70]",

sum(case when sc.score<=70 and sc.score>60 then 1 else 0 end) as "[70-60]",

sum(case when sc.score<=60 and sc.score>0 then 1 else 0 end) as "[60-0]"

from sc left join course

on sc.cid = course.cid

group by sc.cid;

查询各科成绩前三名的记录
大坑比。mysql不能group by 了以后取limit，所以不要想着讨巧了，我快被这一题气死了。思路有两种，第一种比较暴力，计算比自己分数大的记录有几条，如果小于3 就select，因为对前三名来说不会有3个及以上的分数比自己大了，最后再对所有select到的结果按照分数和课程编号排名即可。
select * from sc

where (

select count(*) from sc as a

where sc.cid = a.cid and sc.score

)< 3

order by cid asc, sc.score desc;

第二种比较灵巧一些，用自身左交，但是有点难以理解。
先用自己交自己，条件为a.cid = b.cid and a.score 想要查看完整的表可以

select * from sc a

left join sc b on a.cid = b.cid and a.score

order by a.cid,a.score;

查看，发现结果是47行的一个表，列出了类似 01号课里“30分小于50，也小于70，也小于80，也小于90”“50分小于70，小于80，小于90”.....
所以理论上，对任何一门课来说，分数最高的那三个记录，在这张大表里，通过a.sid和a.cid可以联合确定这个同学的这门课的这个分数究竟比多少个其他记录高/低，
如果这个特定的a.sid和a.cid组合出现在这张表里的次数少于3个，那就意味着这个组合（学号+课号+分数）是这门课里排名前三的。
所以下面这个计算中having count 部分其实count()或者任意其他列都可以，这里制定了一个列只是因为比count()运行速度上更快。

select a.sid,a.cid,a.score from sc a

left join sc b on a.cid = b.cid and a.score

group by a.cid, a.sid

having count(b.cid)<3

order by a.cid;

19.查询每门课程被选修的学生数

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 student.SId,student.Sname

from sc,student

where student.SId=sc.SId  

GROUP BY sc.SId

HAVING count(*)=2；

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

select sc.cid, course.cname, AVG(SC.SCORE) as average from sc, course

where sc.cid = course.cid

group by sc.cid

order by average 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
having也可以用来截取结果表，在这里就先得到平均成绩总表，再截取AVG大于85的即可.

select student.sid, student.sname, AVG(sc.score) as aver from student, sc

where student.sid = sc.sid

group by sc.sid

having aver > 85;

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.查询存在不及格的课程
可以用group by 来取唯一，也可以用distinct

select cid from sc

where score< 60

group by cid;

select DISTINCT sc.CId

from sc

where sc.score <60;

31.查询课程编号为 01 且课程成绩在 80 分及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
用having max()理论上也是对的，但是下面那种按分数排序然后取limit 1的更直观可靠

select student.*, sc.score, sc.cid from student, teacher, course,sc

where teacher.tid = course.tid

and sc.sid = student.sid

and sc.cid = course.cid

and teacher.tname = "张三"

having max(sc.score);

select student.*, sc.score, sc.cid from student, teacher, course,sc

where teacher.tid = course.tid

and sc.sid = student.sid

and sc.cid = course.cid

and teacher.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
为了验证这一题，先修改原始数据

UPDATE sc SET score=90

where sid = "07"

and cid ="02";

这样张三老师教的02号课就有两个学生同时获得90的最高分了。
这道题的思路继续上一题，我们已经查询到了符合限定条件的最高分了，这个时候只用比较这张表，找到全部score等于这个最高分的记录就可，看起来有点繁复。

select student.*, sc.score, sc.cid from student, teacher, course,sc

where teacher.tid = course.tid

and sc.sid = student.sid

and sc.cid = course.cid

and teacher.tname = "张三"

and sc.score = (

    select Max(sc.score)

    from sc,student, teacher, course

    where teacher.tid = course.tid

    and sc.sid = student.sid

    and sc.cid = course.cid

    and teacher.tname = "张三"

);

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
同上，在这里用了inner join后会有概念是重复的记录：“01 课与 03课”=“03 课与 01 课”，所以这里取唯一可以直接用group by

select  a.cid, a.sid,  a.score from sc as a

inner join

sc as b

on a.sid = b.sid

and a.cid != b.cid

and a.score = b.score

group by cid, sid;

36.查询每门功成绩最好的前两名
同上19题

select a.sid,a.cid,a.score from sc as a

left join sc as b

on a.cid = b.cid and a.score

group by a.cid, a.sid

having count(b.cid)<2

order by a.cid;

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

select student.*

from sc ,student

where sc.SId=student.SId

GROUP BY sc.SId

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

40.查询各学生的年龄，只按年份来算
不想做，一般都用41题的方法精确到天

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

select student.SId as 学生编号,student.Sname  as  学生姓名,

TIMESTAMPDIFF(YEAR,student.Sage,CURDATE()) as 学生年龄

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;