MySQL连续案例续集

1、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

分析：平均 avg：GROUP BY分组
从高到低：ORDER BY
所有学生的所有课程的成绩：行转列
所有学生----外联（所有）：RIGHT JOIN右联

语句:
SELECT
	s.sid,
	s.sname ,
	sum((case when sc.cid='01' then sc.score end))语文,
	sum((case when sc.cid='02' then sc.score end))数学,
	sum((case when sc.cid='03' then sc.score end))英语,
   ROUND(avg(sc.score),2) 
FROM
	t_mysql_score sc
	RIGHT JOIN t_mysql_student s ON sc.sid = s.sid 
GROUP BY
	s.sid,
	s.sname

2、检索" 01 "课程分数小于 60，按分数降序排列的学生信息

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

3、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,
	avg(sc.score) 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

4、查询没学过"张三"老师讲授的任一门课程的学生姓名

SELECT
	s.sid,
	s.sname 
FROM
	t_mysql_score sc,
	t_mysql_student s 
WHERE
	s.sid = sc.sid 
	AND 
	sc.cid NOT IN 
	(SELECT cid FROM t_mysql_course c, t_mysql_teacher t 
	WHERE c.tid = t.tid AND t.tname = '张三')
GROUP BY
	s.sid,
	s.sname

5、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_student s
	LEFT JOIN t_mysql_score sc ON s.sid = sc.sid 
GROUP BY
	s.sid,
	s.sname 
HAVING
	n < (SELECT count(*)  FROM t_mysql_course)

6、查询学过「张三」老师授课的同学的信息

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

7、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>=90,1,0))/(SELECT count(1) 
from
		t_mysql_student)*100,2),'%')  优秀率,
		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>=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>=60,1,0))/(SELECT count(1) 
		from t_mysql_student)*100,2),'%') 及格率
		
	FROM
		t_mysql_score sc
		LEFT JOIN t_mysql_course c ON sc.cid = c.cid 
	GROUP BY
		c.cid,
		c.cname