MySQL面试题(下)

09）查询学过「张三」老师授课的同学的信息

SELECT
 
s.*,c.cname,t.tname
 
FROM
 
t_mysql_teacher t,t_mysql_student s,t_mysql_course c,t_mysql_score sc
 
WHERE
 
t.tid=c.tid and c.cid=sc.cid and sc.sid=s.sid and tname = '张三'

10）查询没有学全所有课程的同学的信息

SELECT
 
s.sid,s.sname,count(sc.score) n
 
FROM
 
t_mysql_score sc,t_mysql_student s
 
WHERE
 
sc.sid=s.sid
 
GROUP BY
 
s.sid,s.sname
 
HAVING
 
n<(select count(c.cid) from t_mysql_course c )

11）查询没学过"张三"老师讲授的任一门课程的学生姓名

SELECT s.* FROM t_mysql_student s where s.sid not in(
 
SELECT
 
sc.sid
 
FROM
 
t_mysql_teacher t,t_mysql_course c,t_mysql_score sc
 
WHERE
 
t.tid=c.tid and c.cid=sc.cid and t.tname='张三'
 
GROUP BY
 
sc.sid)

12）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.sid,s.sname,ROUND(AVG(sc.score)) 平均成绩,COUNT(sc.cid) n
 
FROM
 
t_mysql_student s,t_mysql_score sc
 
WHERE
 
s.sid=sc.sid and sc.score<60
 
GROUP BY
 
s.sid,s.sname
 
HAVING
 
n>=2

13）检索" 01 "课程分数小于 60，按分数降序排列的学生信息

SELECT
 
s.*
 
FROM
 
t_mysql_score sc,t_mysql_student s
 
WHERE
 
sc.sid=s.sid and sc.score<60 and cid='01'
 
ORDER BY sc.score DESC

14）按平均成绩从高到低显示所有学生的所有课程的成绩以及平均成绩

SELECT
s.sid,s.sname,round(AVG(sc.score),2) avgNum ,
max(case when sc.cid='01' then sc.score end)语文,
max(case when sc.cid='02' then sc.score end)数学,
max(case when sc.cid='03' then sc.score end)英语
FROM
t_mysql_score sc,t_mysql_student s,t_mysql_course c
WHERE
 sc.sid=s.sid and sc.cid=c.cid
GROUP BY
s.sid,s.sname
ORDER BY avgNum desc
 

15）查询各科成绩最高分、最低分和平均分：
以如下形式显示：课程 ID，课程 name，最高分，最低分，平均分，及格率，中等率，优良率，优秀率及格为>=60，中等为：70-80，优良为：80-90，优秀为：>=90
要求输出课程号和选修人数，查询结果按人数降序排列，若人数相同，按课程号升序排列

SELECT
	c.cid,
	c.cname,
	count(sc.sid) 人数,
	max(sc.score) 最高分,
	min(sc.score) 最低分,
	ROUND(avg(sc.score),2) 平均分,
	CONCAT(ROUND(sum(if(sc.score>=60,1,0))/(SELECT COUNT(1) 
	from t_mysql_student)*100 ,2),'%') 及格率,
	CONCAT(ROUND(sum(if(sc.score>=70 and sc.score<80,1,0))/(SELECT COUNT(1) 
	from t_mysql_student)*100 ,2),'%') 中等率,
	CONCAT(ROUND(sum(if(sc.score>=80 and sc.score<90,1,0))/(SELECT COUNT(1) 
	from t_mysql_student)*100 ,2),'%') 优良率,
	CONCAT(ROUND(sum(if(sc.score>=90,1,0))/(SELECT COUNT(1) 
	from t_mysql_student)*100 ,2),'%') 优秀率
FROM
	t_mysql_score sc
	LEFT JOIN t_mysql_course c ON sc.cid = c.cid 
GROUP BY
	c.cid,
	c.cname