2020-08-17 开始刷第二套题目

练习数据

数据表

–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
创建测试数据

学生表 Student：
导入数据方法：将以下 mysql 语句，完整复制到 workbench 语句窗口（或者是 mysql 的黑窗口），然后运行即可导入，不需要另外创建表，下面表的操作一样。这些语句第一条是创建表（create table），后面都是插入数据到表中（insert into table ）。
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’ , ‘1990-05-20’ , ‘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(‘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’ , ‘2017-12-30’ , ‘女’);
insert into Student values(‘12’ , ‘赵六’ , ‘2017-01-01’ , ‘女’);
insert into Student values(‘13’ , ‘孙七’ , ‘2018-01-01’ , ‘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);

练习题目

– 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息
– 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：
– 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
– 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
– 15.1 按各科成绩进行排序，并显示排名， Score 重复时合并名次
– 16. 查询学生的总成绩，并进行排名，总分重复时保留名次空缺
– 16.1 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

-- 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 s1.*,s2.* from SC s1
join SC s2
on s1.sid=s2.sid
where s1.cid=01 and s2.cid=02
and s1.score>s2.score;

-- 1.1 查询同时存在" 01 "课程和" 02 "课程的情况

select SC1.SID,SC1.CID,SC1.SCORE,SC2.CID,SC2.SCORE from SC SC1
JOIN (select SC2.SID,SC2.CID,SC2.SCORE from SC SC2
WHERE SC2.CID='02') SC2
ON SC1.SID=SC2.SID
WHERE SC1.CID='01';

#answer
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 )
select * from SC SC1
   LEFT join (select * from sc where sc.CId = '02') as SC2
   ON SC1.SID=SC2.SID
where 
SC1.CID='01' ;


-- 1.3 查询不存在" 01 "课程但存在" 02 "课程的情况

select * from SC SC1
   RIGHT join (select * from sc where sc.CId = '02') as SC2
   ON SC1.SID=SC2.SID
where 
SC1.CID NOT IN (01,02) ;
#上面是错的，以下为正确答案
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
select s.sid,s.sname,avg(SC.score) from student s
join SC on s.sid=SC.sid
group by s.sid,s.sname
having avg(SC.score)>=60;


-- 3.	查询在 SC 表存在成绩的学生信息

select Student.* from Student
where Student.sid in (select distinct SC.sid from SC);

-- 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 s.sid,s.sname,count(SC.cid),sum(SC.score)
from student s left join SC
on s.sid=SC.sid
group by s.sid,s.sname;


-- 4.1 查有成绩的学生信息

select s.sid,s.sname,count(SC.cid),sum(SC.score)
from student s left join SC
on s.sid=SC.sid
group by s.sid,s.sname
having count(SC.cid)!=0;

-- 5.	查询「李」姓老师的数量

select count(1) 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 
join sc on sc.sid=student.sid
join course on sc.cid=course.cid
join teacher on Teacher.tid=course.tid
where teacher.tname='张三';

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
#不会
select student.* from student 
where student.sid in 
(select sid from SC group by sid having count(SC.cid) 
!= (select count(distinct 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息

select distinct student.* from student join SC
on student.sid=SC.sid
where SC.cid in (select distinct cid from SC where 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
#没做出来
-- select distinct student.* from student 
-- join SC on student.sid=SC.sid
-- join (select distinct cid from SC where sid='01') s01
-- on s01.cid=SC.cid
-- where count(SC.cid)=(select count(distinct) cid from SC where sid='01');

select * from Student where sid in (
SELECT sid FROM SC WHERE 
cid in (SELECT cid FROM SC WHERE sid=01) AND sid!=01 GROUP BY sid HAVING COUNT(*)=(SELECT COUNT(*) FROM SC WHERE 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

-- select student.sname from student join 	SC on student.sid=SC.sid
-- where SC.cid not in(
-- select SC.cid from SC join course on course.cid=SC.cid
-- join teacher on teacher.tid=course.tid where teacher.tname="张三");

#以上是错的
#思路：找出学过的，用not in
select sid from student where sid  not in( 
select Sid from SC WHERE cid in (
select cid from course join teacher on course.tid=teacher.tid
where 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

select s.sid,s.sname,avg(SC.score) from student s join SC
on s.sid=SC.sid
where s.sid in (select a.sid from (select SC.sid,count(SC.score < 60) from SC
group by SC.sid having count(SC.score < 60) >=2) a )
group by s.sid,s.sname;
#以上是我做的错误的
#这一步就做错了
select SC.sid,count(SC.score < 60) from SC
group by SC.sid having count(SC.score < 60) >=2;


#应该改成这样
select SC.sid,count(SC.score) from SC
where SC.score<60
group by SC.sid ;


#改过后：

select s.sid,s.sname,avg(SC.score) from student s join SC
on s.sid=SC.sid
where s.sid in (select a.sid from (select SC.sid,count(SC.score) from SC
where SC.score<60
group by SC.sid ) a )
group by s.sid,s.sname;


#以下为答案：
select student.sid, student.sname, AVG(sc.score) from student,sc
where 
    student.sid = sc.sid and sc.score<60
group by sc.sid 
having count(*)>1;

-- 12.	检索" 01 "课程分数小于 60，按分数降序排列的学生信息

select distinct student.*,sc.score from student join sc
on sc.sid=student.sid
where sc.cid='01' and sc.score<60
order by sc.score;
#忘了降序
#以下为答案：
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 sid,cid,avg(score) from SC 
group by sid,cid
order by avg(score) desc;
#以上是我做的，没有审好题

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

#只会做到这里，其他的应该是用 case when
select c.cid,c.cname,max(SC.score),min(SC.score),avg(SC.score) from course c
join SC on c.cid=SC.cid
group by c.cid,c.cname;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
#自己本来打算用窗口函数，但是不太熟练还是百度了
select cid,score,rank() over (partition by cid order by score desc) as ranking from SC;
#这是别人的答案1
select a.cid, a.sid, a.score, count(b.score)+1 as ranking
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, ranking ASC;

-- 15.1 按各科成绩进行排序，并显示排名， Score 重复时合并名次
#按cid分
select cid,score,dense_rank() over (partition by cid order by score desc) as ranking from SC;
#不按cid分
select cid,score,dense_rank() over (order by score desc) as ranking from SC;


#答案
select 
sid,cid,score,@rank:=@rank+1 as rn
from 
sc,(select @rank:=0) as t order by score desc;

-- 16.	查询学生的总成绩，并进行排名，总分重复时保留名次空缺
#错了
select sid,sum(score) as total,
rank() over (partition by cid order by total desc) as ranking 
from SC
group by sid
order by ranking;

set @crank=0;
select q.sid, total, @crank := @crank +1 as ranking from(
select sc.sid, sum(sc.score) as total from sc
group by sc.sid
order by total desc)q;

-- 16.1 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

#不会，以下为答案
select s.*,@rank:=if(@sco=scos,"",@rank+1) as rn,
@sco:=scos
from (
select sid,sum(score) as scos
from sc
group by sid 
order by scos desc)s,
(select @rank:=0,@sco:=null) as t;

#看不懂

-- 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

select SC.cid,c.cname,
count(SC.sid) as 人数,
sum(case when sc.score<=100 and sc.score>85 then 1 else 0 end )/count(*)as 八十五到一百,
sum(case when sc.score<=85 and sc.score>70 then 1 else 0 end )/count(*)as 七十到八十五,
sum(case when sc.score<=70 and sc.score>60 then 1 else 0 end )/count(*)as 六十到七十,
sum(case when sc.score<=60 and sc.score>=0 then 1 else 0 end )/count(*)as 零到六十
from SC join Course c
group by SC.cid,c.cname;

#答案：可能业务需要百分比符号，可以用concat拼接
SET sql_mode=(SELECT REPLACE(@@sql_mode,'ONLY_FULL_GROUP_BY',''));
select *,
concat(sum(case when sc.score<=100 and sc.score>85 then 1 else 0 end )/count(*),"%")as 八十五到一百,
concat(sum(case when sc.score<=85 and sc.score>70 then 1 else 0 end )/count(*),"%")as 七十到八十五,
concat(sum(case when sc.score<=70 and sc.score>60 then 1 else 0 end )/count(*),"%")as 六十到七十,
concat(sum(case when sc.score<=60 and sc.score>=0 then 1 else 0 end )/count(*),"%")as 零到六十
from SC 
group by SC.cid;


-- 18.	查询各科成绩前三名的记录

select t.SID,t.CID,t.score,t.ranking FROM 
(select SID,CID,score,dense_rank() over (partition by cid order by score desc) as ranking FROM SC) as t
where t.ranking<=3
order by t.cid;

-- 19.	查询每门课程被选修的学生数
select cid,count(distinct 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

select s.sid,s.sname,count(distinct SC.cid) as countCid from SC join student s
on s.sid=SC.sid
group by s.sid,s.sname
having countCid=2;


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
);

-- 21.	查询男生、女生人数

select sum(case when ssex='男' then 1 else 0 end) as countMale, 
sum(case when ssex='女' then 1 else 0 end) as countFemale from Student;

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 sname like "%风%";

-- 23.	查询同名同性学生名单，并统计同名人数

#不会
select a.*,count(1) as cons
from student a 
inner join Student b
on a.sname = b.sname and a.ssex=b.ssex and a.sid != b.sid;

select * from Student 
left join (select sname,ssex,count(*)同名人数 from Student group by sname,ssex ) as t1
on Student.sname = t1.sname and Student.ssex=t1.ssex
where t1.同名人数>1;

-- 24.	查询 1990 年出生的学生名单

select * from student 
where year(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 s.cid,s.avgScore,dense_rank() over (order by avgScore desc,cid) from(
select cid,avg(score) as avgScore from SC group by cid
) s;

#题目只刷排序没有说要显示ranking
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

select S.sid,avg(SC.score) AS AVGSCORE , S.SNAME FROM SC JOIN
STUDENT S  
ON SC.SID=S.SID
GROUP BY S.SID 
HAVING AVGSCORE >=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 S.SNAME,SC.SCORE FROM SC JOIN STUDENT S
ON SC.SID=S.SID
JOIN COURSE C ON C.CID=SC.CID
WHERE C.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 S.SNAME,SC.CID,SC.SCORE FROM STUDENT S left JOIN SC 
ON SC.SID=S.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 s.sname,c.cname,SC.score from student s 
join SC on SC.sid=s.sid
join course c on c.cid=SC.CId
where SC.sid in (select sid from (select sid,cid,score from SC where score >70)a);

-- 30.	查询不及格的课程

select distinct cid from SC where 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 s.sid,s.sname from student s 
join SC on s.sid=SC.sid
where SC.cid='01' and SC.score>=80;


-- 32.	求每门课程的学生人数

select cid,count(Sid) FROM SC GROUP BY 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

select s.*, SC.score 
from student s 
join SC on s.sid=SC.sid
join Course c on c.cid=SC.cid
join Teacher t on t.tid=C.TId
WHERE T.TNAME="张三"
order by SC.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

#不会

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"
);

#答案:但是报错
select sid,score,cid,tname,rn
from
(

select *, @rank:=if(@score=score,@rank,@rank+1) as rn, @score:=score from 

(select a.sid,a.score,c.cid,d.tname from sc a left join stduent b
on a.sid=b.sid
left join course c 
on a.cid=c.cid
left join teacher d
on c.tid=d.tid
where d.tname="张三"
order by a.score desc)a,

(select @score:=null, @ rank:=0)b

)c
where rn=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
#不会
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

select a.cid,a.sid,a.score,a.ranking
from (select cid,sid,score,dense_rank() over (partition by cid order by score desc) as ranking from SC) as a
where a.ranking <=2;

select cid,sid,score, dense_rank() over (partition by cid order by score desc) as ranking from SC;


select a.sid,a.cid,a.score from sc as a 
left join sc as b 
on a.cid = b.cid and a.score<b.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 cid,count(distinct sid) as countStudent from SC 
group by cid
having countStudent>5;


-- 38.	检索至少选修两门课程的学生学号

select sid,count(distinct cid) as countCourse from SC 
group by sid
having countCourse>=2;

#注意上面应该是>=

-- 39.	查询选修了全部课程的学生信息

select distinct student.* from student 
join SC on student.sid=SC.sid
where student.sid in (


select b.sid from (


select sid,count(cid) from SC group by sid having count(cid)=(select count(distinct cid) from SC))b


);

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

select *,year(now())-year(sage) as age 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
#错了
select *,datediff(day,sage,now()) as age from student;

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;