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# MySQL题目练习及答案（使用SQLyog）

分类: 文章 • 2025-05-08 22:22:46

1.创建表

/*学生表 Student*/
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' , '1990-12-20' , '男');
insert into Student values('04' , '李云' , '1990-12-06' , '男');
insert into Student values('05' , '周梅' , '1991-12-01' , '女');
insert into Student values('06' , '吴兰' , '1992-01-01' , '女');
insert into Student values('07' , '郑竹' , '1989-01-01' , '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' , '2012-06-06' , '女');
insert into Student values('12' , '赵六' , '2013-06-13' , '女');
insert into Student values('13' , '孙七' , '2014-06-01' , '女');

/*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');

/*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');

/*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);

# MySQL题目练习及答案（使用SQLyog）
# MySQL题目练习及答案（使用SQLyog）
# MySQL题目练习及答案（使用SQLyog）
# MySQL题目练习及答案（使用SQLyog）

2. 题目及答案

  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 * 
FROM student RIGHT JOIN 
           (SELECT sc1.SId, sc1.Score AS Score1, sc2.Score AS Score2
            FROM sc sc1 JOIN sc sc2 ON sc1.SId = sc2.SId
            WHERE sc1.CId = "01" AND sc2.CId = "02" AND sc1.Score > sc2.Score) a
 ON a.SId = student.SId

结果：
# MySQL题目练习及答案（使用SQLyog）

  1. 查询同时存在" 01 “课程和” 02 "课程的情况
SELECT sc1.SId, sc1.Score AS Score1, sc2.Score AS Score2
FROM sc sc1 JOIN sc sc2 ON sc1.SId = sc2.SId
WHERE sc1.CId = "01" AND sc2.CId = "02"

# MySQL题目练习及答案（使用SQLyog）

  1. 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 sc.CId = '01') AS sc1 
LEFT JOIN 
    (SELECT * FROM sc WHERE sc.CId = '02') AS sc2
ON sc1.SId = sc2.SId
WHERE sc2.score IS NULL

# MySQL题目练习及答案（使用SQLyog）

  1. 查询不存在" 01 “课程但存在” 02 "课程的情况
SELECT *
FROM 
    (SELECT * FROM sc WHERE sc.CId = '02') AS sc2 
LEFT JOIN 
    (SELECT * FROM sc WHERE sc.CId = '01') AS sc1
ON sc1.SId = sc2.SId
WHERE sc1.score IS NULL

# MySQL题目练习及答案（使用SQLyog）

  1. 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 sc.SId, Sname, SUM(Score)/COUNT(sc.SId) AS Score_Ave
FROM sc JOIN student ON sc.SId = student.SId
GROUP BY sc.SId,Sname
HAVING Score_Ave >= 60

# MySQL题目练习及答案（使用SQLyog）

  1. 查询在 SC 表存在成绩的学生信息
SELECT * 
FROM student
WHERE SId IN (SELECT DISTINCT SId FROM sc)

# MySQL题目练习及答案（使用SQLyog）

  1. 查询所有同学的学生编号、学生姓名、各课程的成绩(没成绩的显示为 null )
SELECT * FROM student 
LEFT JOIN 
(SELECT * FROM sc WHERE sc.CId = "01") sc1 ON student.SId = sc1.SId
LEFT JOIN 
(SELECT * FROM sc WHERE sc.CId = "02") sc2 ON student.SId = sc2.SId
LEFT JOIN
(SELECT * FROM sc WHERE sc.CId = "03") sc3 ON student.SId = sc3.SId

# MySQL题目练习及答案（使用SQLyog）

  1. 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 * 
FROM student 
LEFT JOIN 
(SELECT SId, COUNT(Score), SUM(Score) FROM sc GROUP BY SId) sc1 
ON student.SId = sc1.SId

# MySQL题目练习及答案（使用SQLyog）

  1. 查询「李」姓老师的数量
SELECT SUM(CASE WHEN Tname LIKE "李%" THEN 1 ELSE 0 END) AS "姓李的老师数量"
FROM teacher

# MySQL题目练习及答案（使用SQLyog）

  1. 查询学过「张三」老师授课的同学的信息
SELECT * FROM student
WHERE SId IN (SELECT SId FROM sc 
			  WHERE CId IN (SELECT CId 
			  			   FROM course JOIN teacher 
			  			     		   ON course.TId = teacher.TId 
						   WHERE Tname = "张三"))

# MySQL题目练习及答案（使用SQLyog）

  1. 查询没有学全所有课程的同学的信息
SELECT * FROM student
WHERE SId IN (SELECT SId FROM sc
			  GROUP BY SId
			  HAVING COUNT(CId) = (SELECT COUNT(CId) FROM course))

# MySQL题目练习及答案（使用SQLyog）

  1. 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
WHERE SId IN (SELECT DISTINCT SId FROM sc
			  WHERE CId IN (SELECT CId FROM sc WHERE SId = "01"))

# MySQL题目练习及答案（使用SQLyog）

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