SQL练习题 （学生信息表 教师信息 课程信息等）
2024-05-08 14:20:20
– 创建学生信息表
CREATE TABLE Student(
Sid VARCHAR(10),
Sname VARCHAR(10),
Sage datetime,
Ssex VARCHAR(10)
);
– 由于建表没有定义字符编码 utf8
alter table student default character set utf8;
alter table student change ssex ssex varchar(10) character SET utf8;
SHOW create TABLE Student;
SELECT * FROM student;– 添加数据
> INSERT into Student VALUES('01',"赵雷",'1993-11-01','男'); INSERT into
> Student VALUES('02','钱电','1993-01-01','男'); INSERT into Student
> VALUES('03','孙风','1994-11-21','男'); INSERT into Student
> VALUES('04','李云','1990-03-01','男'); INSERT into Student
> VALUES('05','周梅','1990-10-01','女'); INSERT into Student
> VALUES('06','吴兰','1994-01-31','女'); INSERT into Student
> VALUES('07','郑竹','1990-12-01','女'); INSERT into Student
> VALUES('08','张三','1996-05-01','女'); INSERT into Student
> VALUES('09','李四','1990-01-11','女'); INSERT into Student
> VALUES('10','李四','1990-11-23','女'); INSERT into Student
> VALUES('11','赵六','1994-12-04','女'); INSERT into Student
> VALUES('12','孙七','1991-01-30','男'); INSERT into Student
> VALUES('13','尧舜','1995-01-01','男');
– 科目表
CREATE TABLE Course(
CId VARCHAR(10),
Cname VARCHAR(10),
TId VARCHAR(10)
);
alter table course default character set utf8;
alter table course change cname cname varchar(10) character SET utf8;
INSERT INTO Course VALUES('01','语文','02');
INSERT INTO Course VALUES('02','数学','01');
INSERT INTO Course VALUES('03','英文','03');
– 教师表
CREATE TABLE Teacher(
Tid VARCHAR(10),
Tname VARCHAR(10)
);
alter table teacher change tname tname varchar(10) character SET utf8;
INSERT INTO teacher VALUES('01','张三');
INSERT INTO teacher VALUES('02','李四');
INSERT INTO teacher VALUES('03','王五');
– 成绩表
CREATE TABLE SC(
sid VARCHAR(10),
cid VARCHAR(10),
score DECIMAL(18,1)
);
添加数据
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',90);
– 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
SELECT s2.score FROM student s1 JOIN sc s2 ON s1.sid= s2.sid
WHERE s2.cid=1;
SELECT s2.score FROM student s1 JOIN sc s2 ON s1.sid= s2.sid
WHERE s2.cid=2;
SELECT * FROM
(SELECT s2.sid,s2.score class1 FROM student s1 JOIN sc s2 ON s1.sid= s2.sid
WHERE s2.cid=1) t1,
(SELECT s2.sid,s2.score class2 FROM student s1 JOIN sc s2 ON s1.sid= s2.sid
WHERE s2.cid=2) t2
WHERE t1.sid=t2.sid AND t1.class1>t2.class2;
SELECT * FROM student JOIN (
SELECT t1.sid,class1,class2 FROM
(SELECT s2.sid,s2.score class1 FROM student s1 JOIN sc s2 ON s1.sid= s2.sid
WHERE s2.cid=1) t1,
(SELECT s2.sid,s2.score class2 FROM student s1 JOIN sc s2 ON s1.sid= s2.sid
WHERE s2.cid=2) t2
WHERE t1.sid=t2.sid AND t1.class1>t2.class2
) r ON student.sid=r.sid;
2、查询同时存在" 01 “课程和” 02 "课程的情况
SELECT * FROM sc
WHERE cid=1;
SELECT * FROM sc
WHERE cid=2;
SELECT * FROM (SELECT * FROM sc
WHERE cid=1)c1 JOIN
((SELECT * FROM sc
WHERE cid=2)
)c2 ON c1.sid=c2.sid;
3、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 (SELECT * FROM sc
WHERE cid=1)c1 LEFT JOIN
((SELECT * FROM sc
WHERE cid=2)
)c2 ON c1.sid=c2.sid;
4、查询不存在" 01 “课程但存在” 02 "课程的情况
SELECT sid FROM sc
WHERE cid=1;-- 课程01
SELECT * FROM sc
WHERE sid not in(
SELECT sid FROM sc
WHERE cid=1
) and cid=2;
5、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 sid,AVG(score) 平均成绩 FROM sc
GROUP BY sid
HAVING 平均成绩>60;-- 平均成绩>60
SELECT * FROM student s1 JOIN (SELECT sid,AVG(score) 平均成绩 FROM sc
GROUP BY sid
HAVING 平均成绩>60) s2 ON s1.sid=s2.sid;
6、查询在 SC 表存在成绩的学生信息
SELECT * FROM student s1 RIGHT JOIN sc s2 ON s1.sid=s2.sid;
7、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 sid,SUM(score) FROM sc
GROUP BY sid;-- 总成绩
SELECT sid,COUNT(cid) FROM sc
GROUP BY sid;-- 选课总数
SELECT * FROM student s1 JOIN (SELECT sid,SUM(score) FROM sc
GROUP BY sid) s2 ON s1.Sid=s2.sid JOIN (SELECT sid,COUNT(cid) FROM sc
GROUP BY sid) s3 ON s1.Sid=s3.sid;
8、查有成绩的学生信息
SELECT * FROM student s1 JOIN sc s2 ON s1.sid=s2.sid;
9、查询「李」姓老师的数量
SELECT COUNT(tname) FROM teacher
WHERE tname LIKE '李%';
10、查询学过「张三」老师授课的同学的信息
SELECT tid FROM teacher
WHERE tname LIKE '张三';-- 张三老师授课id
SELECT * from sc
where cid=(SELECT tid FROM teacher
WHERE tname LIKE '张三');-- 选择张三老师的学生id
SELECT * from student s1 JOIN (
SELECT * from sc
where cid=(SELECT tid FROM teacher
WHERE tname LIKE '张三')) s2 ON s1.sid=s2.sid;
11、查询没有学全所有课程的同学的信息
SELECT sid,COUNT(cid) FROM sc
GROUP BY sid
HAVING COUNT(cid)<3; -- 没有学全的学生sid
SELECT * FROM student s1 JOIN (SELECT sid,COUNT(cid) FROM sc
GROUP BY sid
HAVING COUNT(cid)<3) s2 ON s1.Sid=s2.sid;
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 * FROM student s1 RIGHT JOIN sc s2 ON s1.sid=s2.sid;
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 COUNT(cid) FROM sc
WHERE sid=1;-- 学号01选择3门课程
SELECT sid,COUNT(cid)FROM sc
GROUP BY sid
HAVING COUNT(cid)=3;
SELECT * FROM student s1 JOIN (SELECT sid,COUNT(cid)FROM sc
GROUP BY sid
HAVING COUNT(cid)=3) s2 on s1.sid=s2.sid;
14、查询没学过"张三"老师讲授的任一门课程的学生姓名
SELECT tid FROM teacher
WHERE tname LIKE '张三';-- 张三老师授课id
SELECT * from sc
where cid not in (SELECT tid FROM teacher
WHERE tname LIKE '张三');-- 没有选择张三老师的学生id
SELECT * FROM student s1 LEFT JOIN (
SELECT * from sc
where cid not in (SELECT tid FROM teacher
WHERE tname LIKE '张三')) s2 on s1.sid=s2.sid;
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 sid,AVG(score) FROM sc
WHERE score<60
GROUP BY sid;-- 筛选出分数小于60的学生
SELECT *FROM student s1 JOIN (SELECT sid,AVG(score) FROM sc
WHERE score<60
GROUP BY sid) s2 on s1.Sid=s2.sid;
16、检索" 01 "课程分数小于 60，按分数降序排列的学生信息
SELECT * FROM sc
WHERE cid='01'
HAVING score<60
ORDER BY score DESC;-- 符合要求的学生sid
SELECT * FROM student s1 JOIN (SELECT * FROM sc
WHERE cid='01'
HAVING score<60
ORDER BY score DESC) s2 ON s1.Sid=s2.sid;
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 sid,AVG(score) FROM sc
GROUP BY sid
ORDER BY AVG(score) DESC;
SELECT * FROM sc s1 JOIN (SELECT sid,score,AVG(score) avgscore FROM sc
GROUP BY sid) s2 on s1.Sid=s2.sid
ORDER BY avgscore DESC;
18、查询各科成绩最高分、最低分和平均分：
SELECT cid,MAX(score) FROM sc
GROUP BY cid;-- 各科最高分
SELECT cid,MIN(score) FROM sc
GROUP BY cid; -- 各科最低分
SELECT cid,AVG(score) FROM sc
GROUP BY cid;-- 各课成绩平均分
SELECT c1.cid,c1.maxscore,c2.minscore,c3.avgscore FROM
(SELECT cid,MAX(score) maxscore FROM sc
GROUP BY cid) c1 JOIN
(SELECT cid,MIN(score) minscore FROM sc
GROUP BY cid) c2 ON c1.cid=c2.cid JOIN
(SELECT cid,AVG(score) avgscore FROM sc
GROUP BY cid) c3 ON c1.cid=c3.cid;
19、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 cid,
MAX(score) 最高分,
MIN(score) 最低分,
AVG(score) 平均分,
COUNT(*) 选修人数,
SUM(case WHEN score>=60 then 1 ELSE 0 END)/COUNT(*) 及格率,-- 条件字段
SUM(case WHEN score>=70 AND score<80 then 1 ELSE 0 END)/COUNT(*) 中等率,
SUM(case WHEN score>=80 AND score<90 then 1 ELSE 0 END)/COUNT(*) 优良率,
SUM(case WHEN score>=90 then 1 ELSE 0 END)/COUNT(*) 优秀率
FROM sc
GROUP BY cid;
20、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 cid,
MAX(score) 最高分,
MIN(score) 最低分,
AVG(score) 平均分,
COUNT(*) 选修人数,
SUM(case WHEN score>=60 then 1 ELSE 0 END)/COUNT(*) 及格率,-- 条件字段
SUM(case WHEN score>=70 AND score<80 then 1 ELSE 0 END)/COUNT(*) 中等率,
SUM(case WHEN score>=80 AND score<90 then 1 ELSE 0 END)/COUNT(*) 优良率,
SUM(case WHEN score>=90 then 1 ELSE 0 END)/COUNT(*) 优秀率
FROM sc
GROUP BY cid
ORDER BY cid ASC ,COUNT(*) DESC;
21、按各科成绩进行排序，并显示排名， Score 重复时保留名次空缺
SELECT cid,SUM(score) from sc
GROUP BY cid
ORDER BY SUM(score) DESC;
22、按各科成绩进行排序，并显示排名， Score 重复时合并名次
23、查询学生的总成绩，并进行排名，总分重复时保留名次空缺
SELECT sid,SUM(score) FROM sc
GROUP BY sid
ORDER BY SUM(score)DESC;-- 有成绩的学生
SELECT * FROM student s1 LEFT JOIN (SELECT sid,SUM(score) sumscore FROM sc
GROUP BY sid) s2 ON s1.Sid=s2.sid
ORDER BY sumscore DESC;
24、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
SELECT * FROM student s1 JOIN (SELECT sid,SUM(score) sumscore FROM sc
GROUP BY sid) s2 ON s1.Sid=s2.sid
ORDER BY sumscore DESC;
25、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 cid,
COUNT(cid),
sum(CASE WHEN score>=85 AND score<=100 THEN 1 ELSE 0 END )/COUNT(*),
sum(CASE WHEN score>=70 AND score<85 THEN 1 ELSE 0 END )/COUNT(*),
sum(CASE WHEN score>=60 AND score<70 THEN 1 ELSE 0 END )/COUNT(*),
sum(CASE WHEN score>=0 AND score<60 THEN 1 ELSE 0 END )/COUNT(*)
FROM sc
GROUP BY cid；
26、查询各科成绩前三名的记录
SELECT cid,score FROM sc
GROUP BY cid
WHERE cid in (1,2,3)
ORDER BY score DESC
LIMIT 3;
SELECT s1.* FROM sc s1-- 子查询
WHERE (SELECT COUNT(*) FROM sc s2
WHERE s1.cid=s2.cid AND s1.score<s2.score)<3 -- 这个选择前3名
ORDER BY s1.cid ,s1.score DESC;-- 这个纯属排序
27、查询每门课程被选修的学生数
SELECT cid,COUNT(sid) 学生数 FROM sc
GROUP BY cid;
28、查询出只选修两门课程的学生学号和姓名
SELECT sid,COUNT(cid) cidcount FROM sc
GROUP BY sid
HAVING cidcount=2;-- 选择两门课的学生
SELECT * FROM student s1 JOIN (SELECT sid,COUNT(cid) cidcount FROM sc
GROUP BY sid
HAVING cidcount=2) s2 ON s1.Sid=s2.sid;
29、查询男生、女生人数
SELECT ssex,COUNT(ssex)FROM student
GROUP BY ssex;
30、查询名字中含有「风」字的学生信息
SELECT * FROM student
WHERE sname LIKE '%风%';
31、查询同名同性学生名单，并统计同名人数
SELECT sname,COUNT(sname) FROM student
GROUP BY sname
HAVING COUNT(sname)>1;-- 统计出同名同姓人名字
SELECT s1.*,s2.countsname FROM student s1 JOIN (
SELECT sname,COUNT(sname) countsname FROM student
GROUP BY sname
HAVING COUNT(sname)>1) s2 ON s1.sname=s2.sname;
32、查询 1990 年出生的学生名单
SELECT * FROM student
WHERE sage LIKE 1990;
33、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 cid,AVG(score) avgscore FROM sc
GROUP BY cid
ORDER BY avgscore DESC,cid ASC;
34、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 sid,cid,AVG(score) avgscore FROM sc
GROUP BY sid
HAVING avgscore>85;-- 大于85 的学生
SELECT s1.Sid,s1.sname,s2.avgscore FROM student s1 JOIN (
SELECT sid,cid,AVG(score) avgscore FROM sc
GROUP BY sid
HAVING avgscore>85) s2 ON s1.Sid=s2.sid;
35、查询课程名称为「数学」，且分数低于 60 的学生姓名和分数
SELECT * FROM course
WHERE cname LIKE '数学';
SELECT * FROM sc
WHERE score <60;
SELECT s1.*,s2.score,s3.cname FROM student s1 JOIN (SELECT * FROM sc
WHERE score <60) s2 ON s1.Sid=s2.sid JOIN (SELECT * FROM course
WHERE cname LIKE '数学') s3 ON s2.cid=s3.cid;
36、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 s1.*,s2.cid,s2.score FROM student s1
LEFT JOIN sc s2 ON s1.sid=s2.sid;
37、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 * FROM sc
where score >70;
SELECT s1.sname,c.cname,s2.score FROM student s1 JOIN (
SELECT * FROM sc where score >70) s2 ON s1.sid=s2.sid
JOIN course c ON s2.cid=c.cid;
38、查询不及格的课程
SELECT * FROM sc
WHERE score <60;
SELECT c.cname,s.score FROM (SELECT * FROM sc
WHERE score <60) s JOIN course c ON s.cid=c.CId;
39、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 * FROM sc WHERE cid='01' AND score>80;
SELECT * FROM student s1 JOIN (
SELECT * FROM sc WHERE cid='01' AND score>80) s2 ON s1.sid=s2.sid;
40、求每门课程的学生人数
SELECT cid,COUNT(cid) 学生人数 FROM sc
GROUP BY cid;
SELECT c1.*,c2.cname FROM (SELECT cid,COUNT(cid) 学生人数 FROM sc
GROUP BY cid) c1 JOIN course c2 ON c1.cid=c2.CId;
41、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 MAX(c1.score) maxscore FROM (SELECT * FROM teacher
WHERE tname='张三') t JOIN course c ON t.tid=c.tid JOIN (
SELECT * from sc WHERE cid=2) c1 ON c.cid=c1.cid;
SELECT s1.*,s2.maxscore FROM student s1 JOIN (
SELECT c1.sid,MAX(c1.score) maxscore FROM (SELECT * FROM teacher
WHERE tname='张三') t JOIN course c ON t.tid=c.tid JOIN (
SELECT * from sc WHERE cid=2) c1 ON c.cid=c1.cid
) s2 ON s1.Sid=s2.sid;
42、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
UPDATE sc SET score=90 -- 修改学号为7 的学生成绩
where sid = "07"
and cid ="02";
43、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
44、查询每门功成绩最好的前两名
45、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) countsid FROM sc
GROUP BY cid
HAVING countsid>5;
46、检索至少选修两门课程的学生学号
SELECT s1.*,s2.countcid FROM student s1 JOIN (
SELECT sc.sid,COUNT(cid) countcid FROM sc
GROUP BY sid
HAVING countcid>1) s2 ON s1.Sid=s2.sid;
47、查询选修了全部课程的学生信息
SELECT sc.sid,COUNT(cid) countcid FROM sc
GROUP BY sid
HAVING countcid=3;-- 学生信息
SELECT s1.*,s2.countcid FROM student s1 JOIN (
SELECT sc.sid,COUNT(cid) countcid FROM sc
GROUP BY sid
HAVING countcid=3) s2 ON s1.Sid=s2.sid;
48、查询各学生的年龄，只按年份来算
-- CURDATE 当前日期 利用当前日期减去出生日期就是年龄
select student.*,(YEAR(CURDATE())-YEAR(sage))as 年龄 from student;
以上是我自己每天都做一点题然后完成了，有些题我做的可能有些比较繁琐，如果大家有更好的方法也可以一起讨论哈！还有是有些题我看了网上做的但是我还是不太理解就没有写上去，害怕误导大家，如果有做出来的小伙伴们可以留言指导一下我就更好啦！
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