此次为Sql经典50练习题，数据来源如下，只是自己动手操作了一遍。
作者：LSD_Monkey
链接：https://www.jianshu.com/p/476b52ee4f1b
來源：简书
以下为此次需用到的几个表。
1.学生表
Sid学生编号，Sname学生姓名，Sage学生年龄，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' , '1990-12-20' , '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');
2.科目表
Cid课程编号，Cname课程名称，Tid教师编号
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');
3.教师表
Tid教师编号，Tname教师名称
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');
4.成绩表
Sid学生编号，Cid课程编号，score成绩
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);
以上为数据，下面便为问题和我的答案。
1.查询" 01 "课程比" 02 "课程成绩高的学生的信息及课程分数
1.1 查询同时存在" 01 "课程和" 02 "课程的情况
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 )
1.3 查询不存在" 01 "课程但存在" 02 "课程的情况
2.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
3.查询在 SC 表存在成绩的学生信息
4.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 )
4.1 查有成绩的学生信息
5.查询「李」姓老师的数量
6.查询学过「张三」老师授课的同学的信息
7.查询没有学全所有课程的同学的信息
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
9.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
10.查询没学过"张三"老师讲授的任一门课程的学生姓名
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 t1.SId, class1, class2 from
(select SId, score as class1 from sc where sc.CId = '01')as t1,
(select SId, score as class2 from sc where sc.CId = '02')as t2
where t1.SId = t2.SId AND t1.class1 > t2.class2
) as r
on student.SId = r.SId
1.1查询同时存在" 01 "课程和" 02 "课程的情况
select * from
(select * from sc where CId = '01') as t1,
(select * from sc where CId = '02') as t2
where t1.SId = t2.SId
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 )
select * from
(select * from sc where CId = '01') as t1
left join
(select * from sc where CId = '02') as t2
on t1.SId = t2.SId
1.3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 <> '01') as t1
join
(select * from sc where sc.CId = '02') as t2
on t1.SId = t2.SId
2.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 student.SId,Sname,avgscore from
student
right join
(select SId,sum(score)/count(SId) as avgscore from sc
group by sc.SId having avgscore >= 60) as t1
on t1.SId = student.SId
3.查询在 SC 表存在成绩的学生信息
select * from student
right join sc on sc.SId = student.SId
4.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 distinct student.* from student
right join sc on sc.SId = student.SId
5.查询「李」姓老师的数量
select count(Tname) from teacher
where Tname like '李%'
6.查询学过「张三」老师授课的同学的信息
select * from student right join
(select SId from sc
where CId = (select CId from course
where TId = (select TId from teacher
where Tname = '张三'))) as t1
on t1.SId = student.SId
7.查询没有学全所有课程的同学的信息
select * from student right join
(select SId,count(CId) as sumcid from sc
group by SId having sumcid < 3) as t1
on t1.SId = student.SId
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 * from student right join
(select distinct SId from sc
where CId in
(select CId from sc
where SId = '01') and SId <> '01') as t1
on t1.SId = student.SId
9.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 * from student right join
(select distinct SId from sc
where CId =
(select CId from sc
where SId = '01') and SId <> '01') as t1
on t1.SId = student.SId
(这道题貌似错了，就选课相等那个条件不知道该怎么表达，之后想到会改正过来）
10.查询没学过"张三"老师讲授的任一门课程的学生姓名
select * from student
where student.sid not in(
select sc.sid from sc,course,teacher
where sc.cid = course.cid
and course.tid = teacher.tid
and teacher.tname= "张三"
)
