MySQL连续案例续集

01）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_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'

02）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( 1 )

FROM

t_mysql_course)

03）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

复制代码

04）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

复制代码

05）检索" 01 "课程分数小于 60，按分数降序排列的学生信息

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

06）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,

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

复制代码

07）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