MySQL题目示例

1.题目示例

09）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’

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_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)

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.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

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,
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
having n<60

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.*,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

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
方法一：
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 ) 平均值
FROM
t_mysql_score sc
RIGHT JOIN t_mysql_student s ON sc.sid = s.sid
GROUP BY
s.sid,
s.sname

方法二：
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 ) 平均值
FROM
t_mysql_score sc
RIGHT JOIN t_mysql_student s ON sc.sid = s.sid
GROUP BY
s.sid,
s.sname

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,
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,
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