Sql statement exercise collection

Form creation

student table

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');

course table

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');

teacher table

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');

scores table

create table Scores(SId varchar(10),CId varchar(10),score decimal(18,1));
insert into Scores values('01' , '01' , 80);
insert into Scores values('01' , '02' , 90);
insert into Scores values('01' , '03' , 99);
insert into Scores values('02' , '01' , 70);
insert into Scores values('02' , '02' , 60);
insert into Scores values('02' , '03' , 80);
insert into Scores values('03' , '01' , 80);
insert into Scores values('03' , '02' , 80);
insert into Scores values('03' , '03' , 80);
insert into Scores values('04' , '01' , 50);
insert into Scores values('04' , '02' , 30);
insert into Scores values('04' , '03' , 20);
insert into Scores values('05' , '01' , 76);
insert into Scores values('05' , '02' , 87);
insert into Scores values('06' , '01' , 31);
insert into Scores values('06' , '03' , 34);
insert into Scores values('07' , '02' , 89);
insert into Scores values('07' , '03' , 98);

Practice questions

1. Query the information and course scores of students with higher course scores in "01 "Curriculum than" 02"

Idea: First filter the information of the course 01, 02, as T1, T2.

	 将T1,T2按学号SID连接，并筛选出01课程分数大于02课程分数的学生，得到TEM
 	 **将TEM和student进行JOIN，以TEM为主，不然会出现NULL**

SELECT S.SID AS ID , S.SNAME AS NAME ,TEM.CLASS1 AS CLASS1,TEM.CLASS2 AS CLASS2 FROM student AS S RIGHT JOIN 
(SELECT T1.SID AS ID,T1.SCORE AS CLASS1,T2.SCORE AS CLASS2 FROM 
		(SELECT * FROM scores WHERE CID =01)T1,
		(SELECT * FROM scores WHERE CID =02)T2 
		WHERE T1.SID=T2.SID AND T1.SCORE > T2.SCORE
)TEM ON TEM.ID =  S.SID

2. Query the situation that "01" course and "02" course exist at the same time

Idea: First filter the information of the course 01, 02, as T1, T2.

	 将两表sid相同的数据提取出来

SELECT * FROM (		
(SELECT * from scores WHERE scores.CId=01)  T1,
(SELECT * from scores WHERE scores.CId=02)  T2
) WHERE t1.sid =T2.sid

3. Query the situation that "01" course exists but may not exist "02" course (displayed as null if it does not exist)

Idea: null will appear when the join method is used

SELECT * FROM 
(SELECT * from scores WHERE scores.CId=01  )T1
 LEFT JOIN 
(SELECT * from scores WHERE scores.CId=02)  T2 
ON T1.SID = T2.SID

4. Query the student number and student name and average score of students whose average score is greater than or equal to 60 points

Idea: Use the groupby method, AVG function to average, and then use the join method

SELECT S.SID, S.SNAME,TEM.avg_score FROM 
(SELECT SID,AVG(SCORE) as avg_score from scores GROUP BY SId  HAVING avg_score >60)TEM 
 LEFT JOIN student S ON S.SID =TEM.SID 

5. Query the information of at least one class that is the same as the student whose student ID is "01"

Idea: Query the course ID of the 01 student report, query all the students who have the first step result of the course IN, query all the student information

select * from student 
where student.sid in (
    select sc.sid from sc 
    where sc.cid in(
        select sc.cid from sc 
        where sc.sid = '01'
    )
)

6. Query the information of the students who have studied the "Zhang San" teacher

SELECT * from student where student.SId in(
SELECT sid from scores WHERE scores.SId in (
select sid from scores WHERE scores.CId =02)).

7. Query the information of other students who are learning the same courses as the students with number "01"

--  得到学生信息
SELECT * from student where sid in( 
--  课程和01完全相同的学生
select sid from scores where sid not in
--  课程不包含在01的课程里面的学生号码
(select sid from scores where sid !=01 and cid not in 
--  01的所有课程号码
(select cid from scores where sid='01'))) a
--  按照sid分组
group by sid 
--  课程总数和01学生相同的总数筛选
HAVING count(*)=(select count(*) from scores where sid='01') 
--  排除掉01自己
and sid !=01