Mysql 练习题 及 答案

–1.学生表
Student(S,Sname,Sage,Ssex) –S 学生编号,Sname 学生姓名,Sage 出生年月,Ssex 学生性别
–2.课程表
Course(C,Cname,T) –C –课程编号,Cname 课程名称,T 教师编号
–3.教师表
Teacher(T,Tname) –T 教师编号,Tname 教师姓名
–4.成绩表
SC(S,C,score) –S 学生编号,C 课程编号,score 分数
*/
–创建测试数据
create table Student(Svarchar(10),Sname varchar(10),Sage datetime,Ssex nvarchar(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(’08’ , ‘王菊’ , ‘1990-01-20’ , ‘女’)
create table Course(Cvarchar(10),Cname,varchar(10),T 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’)
create table Teacher(T 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’)
create table SC(S varchar(10),Cvarchar(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)

mysql…
–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.score AS’01分数’,c.scoreAS ’02分数’

FROM student a

INNER JOIN sc b

ON a.s=b.s AND b.c=’01’

INNER JOIN sc c

ON a.s=c.s AND c.c=’02’

WHERE b.score>c.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
–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 a.s

,a.sname

,AVG(b.score) AS avgnum

FROM student a

INNER JOIN sc b

ON a.s=b.s

GROUP BY a.s

,a.sname

HAVING AVG(b.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
–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

,a.sname

,COUNT(b.c)

,SUM(b.score)

FROM student a

LEFT JOIN sc b

ON a.s=b.s

GROUP BY 1,2
–6、查询”李”姓老师的数量

SELECT COUNT(1) FROM teacherWHERE tname LIKE ‘李%’
–7、查询学过”张三”老师授课的同学的信息

SELECT a.*

FROM student a

INNER JOIN sc b

ON a.s=b.s

INNER JOIN course c

ON b.c=c.c

INNER JOIN teacher d

ON c.t=d.t

WHERE d.tname=’张三’

GROUP BY 1,2,3,4

–方法2

SELECT a.*

FROM student a

LEFT JOIN (

SELECT a.*

FROM student a

INNER JOIN sc b

ON a.s=b.s

INNER JOIN course c

ON b.c=c.c

INNER JOIN teacher d

ON c.t=d.t

WHERE d.tname=’张三’

GROUP BY 1,2,3,4

)b

ON a.s=b.s

WHERE b.s IS NULL

–8、查询没学过”张三”老师授课的同学的信息

SELECT a.*

FROM student a

LEFT JOIN sc b

ON a.s=b.s

WHERE NOT EXISTS(

SELECT *

FROM course aa

INNER JOIN teacher b

ON aa.t=b.t

INNER JOIN sc c

ON aa.c=c.c

WHERE b.tname=’张三’

AND c.s=a.s

)

GROUP BY 1,2,3,4
–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

INNER JOIN sc b

ON a.s=b.s AND b.c=’01’

INNER JOIN sc c

ON a.s=c.s AND c.c=’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 *

from student a

left join sc b

on a.s=b.s and b.c=’01’

left join sc c

on a.s=c.s and c.c=’02’

where b.c=’01’ and c.c is null
–11、查询没有学全所有课程的同学的信息

SELECT a.*

FROM student a

LEFT JOIN sc b

ON a.s=b.s

LEFT JOIN (SELECT COUNT(1) anumFROM course) c

ON 1=1

GROUP BY 1,2,3,4

HAVING MAX(c.anum)>COUNT(b.c)
–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 a.*

FROM student a

INNER JOIN sc b

ON a.s=b.s

WHERE EXISTS(

SELECT 1 FROM sc WHERE s=’01’ AND c=b.c

)

GROUP BY 1,2,3,4
–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

SELECTa.s,a.sname,a.sage,a.ssex

FROM (SELECT a.*,COUNT(b.c) ASsnum

FROM student a

INNER JOIN sc b

ON a.s=b.s

WHERE EXISTS(

SELECT 1 FROM scWHERE s=’01’ AND c=b.c

)

GROUP BY 1,2,3,4)a

INNER JOIN (SELECTa.*,COUNT(b.c) AS anum

FROM student a

INNER JOIN sc b

ON a.s=b.s

GROUP BY 1,2,3,4)b

ON a.s=b.s

INNER JOIN (SELECT COUNT(1) ASnum1 FROM sc WHERE s=’01’)c

ON 1=1

WHERE a.snum=b.anum ANDa.snum=c.num1

–方法二

SELECT a.*

,COUNT(b.c) AS anum

,SUM(CASE WHEN EXISTS(SELECT 1 FROM sc WHERE s=’01’ AND c=b.c)THEN 1 ELSE 0 END) AS snum

,MAX(c.num1) AS num1

FROM student a

INNER JOIN sc b

ON a.s=b.s

INNER JOIN (SELECT COUNT(1) ASnum1 FROM sc WHERE s=’01’)c

ON 1=1

GROUP BY 1,2,3,4

HAVING anum=snum AND anum=num1
–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.*

FROM student a

LEFT JOIN(

SELECT a.s

FROM student a

LEFT JOIN sc b

ON a.s=b.s

LEFT JOIN course c

ON b.c=c.c

LEFT JOIN teacher d

ON c.t=d.t AND d.tname=’张三’

WHERE d.t IS NOT NULL

)b

ON a.s=b.s

WHERE b.s IS NULL
–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

,a.sname

,AVG(b.score)

FROM student a

INNER JOIN sc b

ON a.s=b.s

GROUP BY 1,2

HAVING SUM(CASE WHENb.score>=60 THEN 0 ELSE 1 END)>=2
–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.score

FROM student a

INNER JOIN sc b

ON a.s=b.s

WHERE

b.c=’01’

AND b.score<60

ORDER BY b.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.*

,SUM(CASE WHEN b.c=’01’ THEN b.scoreELSE 0 END) AS s01

,SUM(CASE WHEN b.c=’02’ THEN b.scoreELSE 0 END) AS s02

,SUM(CASE WHEN b.c=’03’ THEN b.score ELSE0 END) AS s03

,AVG(CASE WHEN b.score IS NULL THEN 0ELSE b.score END) avs

FROM student a

LEFT JOIN sc b

ON a.s=b.s

GROUP BY 1,2,3,4

ORDER BY avs 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

,a.cname

,MAX(b.score)

,MIN(b.score)

,AVG(b.score)

,SUM(CASE WHEN b.score>=60 THEN 1ELSE 0 END)/COUNT(1)

,SUM(CASE WHEN b.score>=70 ANDb.score<80 THEN 1 ELSE 0 END)/COUNT(1)

,SUM(CASE WHEN b.score>=80 ANDb.score<90 THEN 1 ELSE 0 END)/COUNT(1)

,SUM(CASE WHEN b.score>=90 THEN 1ELSE 0 END)/COUNT(1)

FROM course a

INNER JOIN sc b

ON a.c=b.c

GROUP BY 1,2
–19、按各科成绩进行排序,并显示排名

SET @rn:=0;

SELECT a.*,@rn:=@rn+1

FROM(

SELECT a.*,b.score

FROM course a

INNER JOIN sc b

ON a.c=b.c

WHERE a.c=’01’

ORDER BY b.c,score DESC

)a
–20、查询学生的总成绩并进行排名

SELECT a.*

,COUNT(b.c)+1 asall

FROM sc a

LEFT JOIN sc b

ON a.c=b.c ANDa.score<b.score

GROUP BY 1,2,3

ORDER BY a.c,asall
–21、查询不同老师所教不同课程平均分从高到低显示

SELECT a.*

,b.cname

,AVG(c.score) ascore

FROM teacher a

INNER JOIN course b

ON a.t=b.t

INNER JOIN sc c

ON b.c=c.c

GROUP BY 1,2,3

ORDER BY ascore 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 a.*,b.cname,b.score

FROM student a

INNER JOIN (

SELECT a.*

,c.cname

,COUNT(b.c)+1 AStp

FROM sc a

LEFT JOIN sc b

ON a.c=b.c AND a.score<b.score

LEFT JOIN course c

ON a.c=c.c

GROUP BY 1,2,3,4

HAVING COUNT(b.c)+1 IN(2,3)

ORDER BY a.c,tp)b

ON a.s=b.s
–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 a.c

,a.cname

,SUM(CASE WHEN b.score<=100 ANDb.score>85 THEN 1 ELSE 0 END) AS ‘[100-85]’

,SUM(CASE WHEN b.score<=85 ANDb.score>70 THEN 1 ELSE 0 END) AS ‘[85-70]’

,SUM(CASE WHEN b.score<=70 ANDb.score>60 THEN 1 ELSE 0 END) AS ‘[70-60]’

,SUM(CASE WHEN b.score<=60 ANDb.score>0 THEN 1 ELSE 0 END) AS ‘[60-0]’

,SUM(CASE WHEN b.score<=100 ANDb.score>85 THEN 1 ELSE 0 END)/COUNT(1) AS ‘[100-85]%’

,SUM(CASE WHEN b.score<=85 ANDb.score>70 THEN 1 ELSE 0 END)/COUNT(1) AS ‘[85-70]%’

,SUM(CASE WHEN b.score<=70 ANDb.score>60 THEN 1 ELSE 0 END)/COUNT(1) AS ‘[70-60]%’

,SUM(CASE WHEN b.score<=60 ANDb.score>0 THEN 1 ELSE 0 END)/COUNT(1) AS ‘[60-0]%’

FROM course a

INNER JOIN sc b

ON a.c=b.c

GROUP BY 1,2
–24、查询学生平均成绩及其名次

SELECT a.*

,COUNT(b.s)+1

FROM (

SELECT a.*,AVG(CASE WHEN b.score IS NULL THEN 0 ELSE b.scoreEND) AS ascore

FROM student a

LEFT JOIN sc b

ON a.s=b.s

GROUP BY 1,2,3,4

)a

LEFT JOIN(

SELECT a.*,AVG(CASE WHEN b.score IS NULL THEN 0 ELSE b.scoreEND) AS ascore

FROM student a

LEFT JOIN sc b

ON a.s=b.s

GROUP BY 1,2,3,4

)b

ON a.ascore<b.ascore

GROUP BY 1,2,3,4,5
–25、查询各科成绩前三名的记录

SELECT a.*,COUNT(b.c)+1 ASascore

FROM sc a

LEFT JOIN sc b

ON a.c=b.c ANDa.score<b.score

GROUP BY 1,2,3

HAVING ascore<=3

ORDER BY a.c,ascore
–26、查询每门课程被选修的学生数

SELECT a.*

,COUNT(b.s)

FROM course a

LEFT JOIN sc b

ON a.c=b.c

GROUP BY 1,2,3
–27、查询出只有两门课程的全部学生的学号和姓名

SELECT a.*

,COUNT(b.c)

FROM student a

LEFT JOIN sc b

ON a.s=b.s

GROUP BY 1,2,3,4

HAVING COUNT(b.c)=2
–28、查询男生、女生人数
–29、查询名字中含有”风”字的学生信息
–30、查询同名同性学生名单,并统计同名人数

SELECT sname

,ssex

,COUNT(1)

FROM student

GROUP BY 1,2

HAVING COUNT(1)>1
–31、查询1990年出生的学生名单(注:Student表中Sage列的类型是datetime)

SELECT * FROM student WHEREYEAR(sage)=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

SELECT a.*

,AVG(b.score) ascore

FROM course a

LEFT JOIN sc b

ON a.c=b.c

GROUP BY 1,2,3

ORDER BYascore DESC,a.c

–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.*

,AVG(b.score) ascore

FROM student a

LEFT JOIN sc b

ON a.s=b.s

GROUP BY 1,2,3,4

HAVING ascore>=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 c.*

FROM course a

LEFT JOIN sc b

ON a.c=b.c

LEFT JOIN student c

ON b.s=c.s

WHERE a.cname=’数学’

AND b.score<60
–35、查询所有学生的课程及分数情况;

SELECT *

FROM sc a

INNER JOIN student b

ON a.s=b.s

INNER JOIN course c

ON a.c=c.c
–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 b.sname

,c.cname

,a.score

FROM sc a

INNER JOIN student b

ON a.s=b.s

INNER JOIN course c

ON a.c=c.c

WHERE a.score>70
–37、查询不及格的课程
–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.*

FROM student a

INNER JOIN sc b

ON a.s=b.s

WHERE b.c=’01’ AND b.score>80
–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,成绩最高的学生信息及其成绩

SELECT a.*,b.score

FROM student a

INNER JOIN sc b

ON a.s=b.s

INNER JOIN(

SELECT c.c

,MAX(c.score) ASmaxscore

FROM teacher a

INNER JOIN course b

ON a.t=b.t

INNER JOIN sc c

ON b.c=c.c

WHERE a.tname=’张三’

GROUP BY c)c

ON b.c=c.c ANDb.score=c.maxscore
–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 a.s

,a.c

,a.score

FROM sc a

INNER JOIN (

SELECT a.score

,b.s

,COUNT(1)

FROM sc a

INNER JOIN student b

ON a.s=b.s

GROUP BY a.score,b.s

HAVING COUNT(1)>1

)b

ON a.s=b.s AND a.score=b.score

方法二

SELECT a.*

FROM sc a

LEFT JOIN sc b

ON a.s=b.s AND a.score=b.scoreAND a.c<>b.c

GROUP BY 1,2,3

HAVING COUNT(b.s)>0
–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

SELECT a.c

,COUNT(1) AS pnum

FROM sc a

GROUP BY 1

HAVING pnum>5

ORDER BY pnum DESC,a.c
–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

SELECT a.*,YEAR(CURDATE())-YEAR(a.sage)

FROM student a
–47、查询本周过生日的学生

SELECT a.*

FROM student a

WHERECONCAT(MONTH(a.sage),DAY(a.sage))>=CONCAT(MONTH(SUBDATE(CURDATE(),WEEKDAY(CURDATE()))),DAY(SUBDATE(CURDATE(),WEEKDAY(CURDATE()))))

AND CONCAT(MONTH(a.sage),DAY(a.sage))<=CONCAT(MONTH(SUBDATE(CURDATE(),WEEKDAY(CURDATE())-6)),DAY(SUBDATE(CURDATE(),WEEKDAY(CURDATE())-6)))

–48、查询下周过生日的学生

–49、查询本月过生日的学生

SELECT a.* FROM student a WHEREMONTH(a.sage)=MONTH(CURDATE())
–50、查询下月过生日的学生

SELECT a.*FROM student a WHERE MONTH(a.sage)=MONTH(CURDATE())+1

有些没有答案的,是因为和上面的差不多,偷个懒。。。。

    原文作者:郝新宝
    原文地址: https://zhuanlan.zhihu.com/p/31093190
    本文转自网络文章,转载此文章仅为分享知识,如有侵权,请联系博主进行删除。
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