Mysql 练习题 及 答案

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…

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 –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(
S varchar(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 SC(
S varchar(10),
C 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);
create table Course(
C varchar(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');

–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.score AS '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

–6、查询姓老师的数量 

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 NOT 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 

SELECT a.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 0 ELSE 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 1 ELSE 0 END)/COUNT(1)
       ,SUM(CASE WHEN b.score>=70 AND b.score<80 THEN 1 ELSE 0 END)/COUNT(1)
       ,SUM(CASE WHEN b.score>=80 AND b.score<90 THEN 1 ELSE 0 END)/COUNT(1)
       ,SUM(CASE WHEN b.score>=90 THEN 1 ELSE 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 AND a.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 AND b.score>85 THEN 1 ELSE 0 END) AS '[100-85]'
       ,SUM(CASE WHEN b.score<=85  AND b.score>70 THEN 1 ELSE 0 END) AS '[85-70]'
       ,SUM(CASE WHEN b.score<=70  AND b.score>60 THEN 1 ELSE 0 END) AS '[70-60]'
       ,SUM(CASE WHEN b.score<=60  AND b.score>0 THEN 1 ELSE 0 END) AS '[60-0]'   
       ,SUM(CASE WHEN b.score<=100 AND b.score>85 THEN 1 ELSE 0 END)/COUNT(1) AS '[100-85]%'
       ,SUM(CASE WHEN b.score<=85  AND b.score>70 THEN 1 ELSE 0 END)/COUNT(1) AS '[85-70]%'
       ,SUM(CASE WHEN b.score<=70  AND b.score>60 THEN 1 ELSE 0 END)/COUNT(1) AS '[70-60]%'
       ,SUM(CASE WHEN b.score<=60  AND b.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 AND a.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 WHERE YEAR(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 AND b.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.score AND 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
WHERE CONCAT(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 WHERE MONTH(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。。。。 

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