mysql 15道语句练习题

表名和字段
–1.学生表
Student(s_id,s_name,s_birth,s_sex) --学生编号,学生姓名, 出生年月,学生性别
–2.课程表
Course(c_id,c_name,t_id) – --课程编号, 课程名称, 教师编号
–3.教师表
Teacher(t_id,t_name) --教师编号,教师姓名
–4.成绩表
Score(s_id,c_id,s_score) --学生编号,课程编号,分数

ps：插入数据就不演示了

练习题及语句

-- 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 S.s_id AS 学生编号,S.s_name AS 学生姓名, A.s_score AS 分数
FROM Student AS S
INNER JOIN Score AS A
ON S.s_id=A.s_id and A.c_id='01'
LEFT JOIN Score as C 
ON C.s_id=S.s_id and C.c_id='02'
WHERE A.c_id>C.c_id;

-- 2、查询"01"课程比"02"课程成绩低的学生的信息及课程分数
SELECT S.s_id AS 学生编号,S.s_name AS 学生姓名, A.s_score AS 分数
FROM Student AS S
INNER JOIN Score AS A
ON S.s_id=A.s_id and A.c_id='01'
LEFT JOIN Score as C 
ON C.s_id=S.s_id and C.c_id='02'
WHERE A.c_id<C.c_id;

-- 3、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 S.s-id AS 学生编号,S.s_name AS 学生姓名,AVG(C.s_score) AS 平均成绩
FROM Student AS S
INNER JOIN Score AS C 
ON S.s_id=C.s_id
GROUP BY S.s_id,S.s_name,C.s_id
HAVING AVG(C.s_score)>=60;

-- 4、查询平均成绩小于60分的同学的学生编号和学生姓名和平均成绩
        -- (包括有成绩的和无成绩的)

select b.s_id,b.s_name,ROUND(AVG(a.s_score),2) as avg_score from 
    student b 
    left join score a on b.s_id = a.s_id
    GROUP BY b.s_id,b.s_name HAVING avg_score <60
    union
select a.s_id,a.s_name,0 as avg_score from 
    student a 
    where a.s_id not in (
    select distinct s_id from score);

-- 5、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

SELECT S.s_id AS 学生编号,S.s_name AS 学生姓名,COUNT(C,c_id) AS 选课总数,SUM(C.s_score) AS 总成绩
FROM Student AS S
INNER JOIN Score AS C ON S.s_id=C.s_id
GROUP BY S.s_id,S.s_name ;

-- 6、查询"李"姓老师的数量 
select count(t_id) from teacher where t_name like '李%';

-- 7、查询学过"张三"老师授课的同学的信息 
SELECT S.s_id AS 学生编号, S.s_name AS 学生姓名,S.s_sex AS 性别 
FROM Student AS S WHERE S.s_id  IN
  (SELECT G.s_id FROM Score AS G
   WHERE G.s_id=S.s_id AND G.c_id IN
   (SELECT C.c_id FROM Couse AS C
   WHERE C.c_id=G.c_id AND C.t_id IN 
(SELECT T.t_id FROM Teacher AS T WHERE T.t-id=C.t_id AND T.t_name = '张三'))); 

-- 8、查询没学过"张三"老师授课的同学的信息 

SELECT S.s_id AS 学生编号, S.s_name AS 学生姓名,S.s_sex AS 性别 
FROM Student AS S WHERE S.s_id  IN
  (SELECT G.s_id FROM Score AS G
   WHERE G.s_id=S.s_id AND G.c_id IN
   (SELECT C.c_id FROM Couse AS C
   WHERE C.c_id=G.c_id AND C.t_id IN 
(SELECT T.t_id FROM Teacher AS T WHERE T.t-id=C.t_id AND T.t_name ! = '张三'))); 

-- 9、查询学过编号为"01"并且也学过编号为"02"的课程的同学的信息

Select *
From student
Where s_id=(Select s_id
From score as s01 ,score as s02
Where s01.s_score=’01’and s02.s_score=’02’
)


-- 10、查询学过编号为"01"但是没有学过编号为"02"的课程的同学的信息
Select *
From student AS s
INNER JOIN Score as a
On s.s_id =a.s_id
Group by s.c_id
Having (select s_id From Score  Where c_id =’01’)!=(select s_id
From score
Where c_id =’02’);

-- 11、查询没有学全所有课程的同学的信息
Select s_id,s_name,s_birth ,s_sex 
From Student
where s_id not in (
select s_id    
from Score
where c_id = ‘01’) or
s_id not in(
select s_id
from Score
where c_id = ‘02’);

-- 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
Where s_id=(
Select s_id
From score
Where c_id on(
Select c_id
From score
Where s_id=’01’
))

-- 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 *
From student
Where s_id=(
Select s_id
From score
Where c_id =(
Select c_id
From score
Where s_id=’01’
))

-- 14、查询没学过"张三"老师讲授的任一门课程的学生姓名 

SELECT S.s_name AS 学生姓名 
FROM Student AS S WHERE S.s_id NOT IN
  (SELECT G.s_id FROM Score AS G
   WHERE G.s_id=S.s_id AND G.c_id IN
   (SELECT C.c_id FROM Couse AS C
   WHERE C.c_id=G.c_id AND C.t_id IN 
(SELECT T.t_id FROM Teacher AS T WHERE T.t-id=C.t_id AND T.t_name = '张三')));

-- 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 s.s_id as 学号,s.s_name as学生姓名 AVG（a.s_score）
FROM Score AS a,Student as s
GROUP BY s_id
HAVING  COUNT(c_id)>=2 AND a.s_core<60

还有什么不足的地方请大佬赐教！！！