SQL练习

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最新推荐文章于 2025-01-11 20:20:55 发布

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文章标签： sql 数据库 java

本文链接：https://blog.youkuaiyun.com/qq_41025577/article/details/128861727

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该文提供了一系列SQL练习题，涉及对学生表、课程表、教师表和成绩表的数据查询，包括比较不同课程成绩、查询特定课程情况、平均成绩筛选、教师数量统计以及学生选课情况等复杂查询操作。

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SQL练习

一.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

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');

课程表 Course(CId,Cname,TId) – CId 课程编号,Cname 课程名称,TId 教师编号

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(TId,Tname)–TId 教师编号,Tname 教师姓名

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(SId,CId,score)–SId 学生编号,CId 课程编号,score 分数

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);

二.题目

题目参考：
http://t.csdn.cn/BGunH
经典50道SQL练习题

三.答案

查询" 01 “课程比” 02 "课程成绩高的学生的信息及课程分数

SELECT t1.sid,t1.score as class1, t2.score as class2,student.*FROM
(SELECT * FROM sc WHERE cid='01') t1,
(SELECT * FROM sc WHERE cid='02') t2,student
WHERE t1.sid=t2.sid AND  t1.score>t2.score  AND t1.sid=student.SId

1.1 查询同时存在" 01 “课程和” 02 "课程的情况

SELECT t1.sid,t1.score as class1, t2.score as class2,student.*FROM
(SELECT * FROM sc WHERE cid='01') t1,
(SELECT * FROM sc WHERE cid='02') t2,student
WHERE t1.sid=t2.sid AND t1.sid=student.SId

SELECT student.* FROM sc,student
WHERE sc.CId='01' AND sc.SId=student.SId in (select sid from sc where cid = '02')

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 )

参考图片* 不等于的三种表示法： <>,not in ,!=，其中not in 的查找结果不能为null，否则查找失败。

//左连接
SELECT * FROM 
(SELECT * FROM sc WHERE sc.CId= '01')t1 
LEFT JOIN
(SELECT * FROM sc WHERE sc.CId != '02')t2
On t1.sid = t2.SId
//右连接
SELECT*FROM
(SELECT* FROM sc WHERE sc.CId != '02')t1
RIGHT JOIN
(SELECT* FROM sc WHERE sc.CId = '01')t2
on t1.sid=t2.SId

 
SELECT *FROM sc
WHERE cid='01' and cid not in(02)

1.3查询不存在" 01 “课程但存在” 02 "课程的情况

SELECT *FROM sc
WHERE cid='02' and cid not in(01)

 
SELECT * FROM 
(SELECT * FROM sc WHERE sc.CId= '02')t1 
LEFT JOIN
(SELECT * FROM sc WHERE sc.CId != '01')t2
On t1.sid = t2.SId

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

//分组查询
SELECT sc.SId,st.Sname,AVG(sc.score) avg_score FROM sc
join student st on sc.SId=st.SId
GROUP BY sc.SId
HAVING avg_score >=60

查询在 SC 表存在成绩的学生信息

//去重查询 DISTINCT 
SELECT DISTINCT sc.SId,student.* FROM sc,student
WHERE student.SId=sc.SId

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 student.Sname,sc.SId,SUM(sc.score) sum_score,COUNT(sc.CId) count_cid FROM sc
LEFT JOIN student
ON sc.SId=student.SId
GROUP BY student.SId

4.1查有成绩的学生信息

SELECT student.* FROM student
LEFT JOIN sc
ON sc.SId=student.SId
GROUP BY student.SId

Select st.* from student st
Where st.sid in (select sid from sc)

查询「李」姓老师的数量

SELECT COUNT(tid) FROM teacher
WHERE tname LIKE '李%'

查询学过「张三」老师授课的同学的信息

SELECT sc.SId,st.* FROM sc
LEFT JOIN student st
on st.SId=sc.SId
WHERE sc.CId in (SELECT c.cid FROM course c
JOIN teacher t
ON c.tid=t.tid
WHERE t.tname= '张三')

查询没有学全所有课程的同学的信息

SELECT *FROM student 
WHERE SId in (		//当sid满足下列条件
	SELECT sid from sc
	GROUP BY sid HAVING COUNT(cid)<(		//根据sid分类，计算cid的数量
		SELECT COUNT(DISTINCT cid) from sc)	//去重复，计算cid的总数
)

SELECT *FROM student
WHERE sid in (		
	SELECT sid FROM sc 		
	GROUP BY SId HAVING COUNT(cid)<	
	(SELECT COUNT(cid)from course)		//cid总数直接去课表里获取
	)

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 sc.sid, st.* from sc
Join student st
On sc.sid = st.sid
Where sc.cid in (Select cid from sc where sid = '01')

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

// 1. 找01所学的课，且排除01 （这一步骤完，数据会有重复），这一步也可以不用
// 2.根据sid分组查找，找cid=3
SELECT * FROM student st
JOIN sc ON st.SId= sc.SId
WHERE sc.CId in (SELECT cid FROM sc where sid ='01') and sc.sid !='01'
GROUP BY sc.SId 
HAVING COUNT(sc.CId)=(
	Select count(cid) from sc where sid = '01')

SELECT * FROM student st
JOIN sc ON st.SId= sc.SId
GROUP BY sc.SId 
HAVING COUNT(sc.CId)=(
	Select count(cid) from sc where sid = '01')

查询没学过"张三"老师讲授的任一门课程的学生姓名

SELECT st.Sname FROM student st,sc  
WHERE st.SId =sc.SId AND sc.SId NOT IN 
	(SELECT SId FROM sc 
	WHERE CId IN 
		(SELECT CId FROM course c,teacher t
		 WHERE c.TId = t.TId AND t.Tname = '张三') 
	 GROUP BY SId) 
GROUP BY sc.SId