目录
一.案例(接上篇博客)
09)查询学过「张三」老师授课的同学的信息
10)查询没有学全所有课程的同学的信息
11)查询没学过"张三"老师讲授的任一门课程的学生姓名
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
13)检索" 01 "课程分数小于 60,按分数降序排列的学生信息
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
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
二.思维导图
SELECT
s.*,
c.cname,
t.tname,
sc.score
FROM
t_mysql_teacher t,
t_mysql_course c,
t_mysql_student s,
t_mysql_score sc
WHERE
t.tid = c.tid
AND c.cid = sc.cid
AND sc.sid = s.sid
AND t.tname = '张三'
-- 没有学全
学全了有多少门:统计一共有多少门学科
统计每一个学生学了多少门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( 1 ) FROM t_mysql_course)
没学过: 子查询
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
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
SELECT
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
平均 GROUP BY
从高到低 ORDER BY
所有学生的所有课程的成绩
姓名 语文 数学 英语 平均成绩select
s.sid,
s.sname,
sum(if(sc.cid = '01',sc.score,0)) 语文,
sum(if(sc.cid = '02',sc.score,0)) 数学,
sum(if(sc.cid = '03',sc.score,0)) 英语,
ROUND(AVG(sc.score),2) 平均分
from
t_mysql_score sc
RIGHT JOIN t_mysql_student s on sc.sid = s.sid
GROUP BY
s.sid,
s.sname
-- GROUP BY
-- 大量使用函数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,
CONCAT(ROUND(sum(if(sc.score >= 80 and sc.score < 90,1,0))/(select count(1) from t_mysql_student)*100,2),'%') 优良,
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