sql練習

一對多的狀況下，不能使用not in .否則的話，剔除了自己，還有不少重複項

 

–1.學生表
Student(s_id,s_name,s_birth,s_sex) --學生編號,學生姓名, 出生年月,學生性別
–2.課程表
Course(c_id,c_name,t_id) – --課程編號, 課程名稱, 教師編號
–3.教師表
Teacher(t_id,t_name) --教師編號,教師姓名
–4.成績表
Score(s_id,c_id,s_score) --學生編號,課程編號,分數

--建表
--學生表
CREATE TABLE `Student`(
    `s_id` VARCHAR(20),
    `s_name` VARCHAR(20) NOT NULL DEFAULT '',
    `s_birth` VARCHAR(20) NOT NULL DEFAULT '',
    `s_sex` VARCHAR(10) NOT NULL DEFAULT '',
    PRIMARY KEY(`s_id`)
);
--課程表
CREATE TABLE `Course`(
    `c_id`  VARCHAR(20),
    `c_name` VARCHAR(20) NOT NULL DEFAULT '',
    `t_id` VARCHAR(20) NOT NULL,
    PRIMARY KEY(`c_id`)
);
--教師表
CREATE TABLE `Teacher`(
    `t_id` VARCHAR(20),
    `t_name` VARCHAR(20) NOT NULL DEFAULT '',
    PRIMARY KEY(`t_id`)
);
--成績表
CREATE TABLE `Score`(
    `s_id` VARCHAR(20),
    `c_id`  VARCHAR(20),
    `s_score` INT(3),
    PRIMARY KEY(`s_id`,`c_id`)
);
--插入學生表測試數據
insert into Student values('01' , '趙雷' , '1990-01-01' , '男');
insert into Student values('02' , '錢電' , '1990-12-21' , '男');
insert into Student values('03' , '孫風' , '1990-05-20' , '男');
insert into Student values('04' , '李雲' , '1990-08-06' , '男');
insert into Student values('05' , '周梅' , '1991-12-01' , '女');
insert into Student values('06' , '吳蘭' , '1992-03-01' , '女');
insert into Student values('07' , '鄭竹' , '1989-07-01' , '女');
insert into Student values('08' , '王菊' , '1990-01-20' , '女');
--課程表測試數據
insert into Course values('01' , '語文' , '02');
insert into Course values('02' , '數學' , '01');
insert into Course values('03' , '英語' , '03');

--教師表測試數據
insert into Teacher values('01' , '張三');
insert into Teacher values('02' , '李四');
insert into Teacher values('03' , '王五');

--成績表測試數據
insert into Score values('01' , '01' , 80);
insert into Score values('01' , '02' , 90);
insert into Score values('01' , '03' , 99);
insert into Score values('02' , '01' , 70);
insert into Score values('02' , '02' , 60);
insert into Score values('02' , '03' , 80);
insert into Score values('03' , '01' , 80);
insert into Score values('03' , '02' , 80);
insert into Score values('03' , '03' , 80);
insert into Score values('04' , '01' , 50);
insert into Score values('04' , '02' , 30);
insert into Score values('04' , '03' , 20);
insert into Score values('05' , '01' , 76);
insert into Score values('05' , '02' , 87);
insert into Score values('06' , '01' , 31);
insert into Score values('06' , '03' , 34);
insert into Score values('07' , '02' , 89);
insert into Score values('07' , '03' , 98);

 

練習題和sql語句

#一、查詢"01"課程比"02"課程成績高的學生的信息及課程分數

select a.* ,b.s_score as 01_score,c.s_score as 02_score 
from student a 
join score b on a.s_id=b.s_id and b.c_id='01' 
left join score c on a.s_id = c.s_id and c.c_id = '02'
where b.s_score > c.s_score;

或者：

select a.*,b.s_score as 01_score,c.s_score as 02_score 
from student a,score b,score c 
where a.s_id=b.s_id 
and a.s_id=c.s_id 
and b.c_id='01' 
and c.c_id='02' 
and b.s_score>c.s_score

 

#查詢平均成績小於60分的同窗和沒有成績的的學生編號和學生姓名和平均成績

select a.s_id,a.s_name,round(avg(b.s_score),2) as avg_score 
from student a
left join score b on a.s_id = b.s_id
group by a.s_id,a.s_name
having avg_score < 60 or avg_score is null

 

 

 

 

 

 

#查詢平均成績大於等於60分的同窗的學生編號和學生姓名和平均成績

select a.s_id,a.s_name,round(avg(b.s_score),2) as avg_score 
from student a
left join score b on a.s_id = b.s_id
group by a.s_id,a.s_name
having avg_score >=60

 

#查詢全部同窗的學生編號、學生姓名、選課總數、全部課程的總成績

select a.s_id,a.s_name,count(b.c_id) as total_course,sum(b.s_score) as total_score
from student a
left join score b on a.s_id = b.s_id
group by a.s_id,a.s_name

 

#查詢"李"姓老師的數量

#

 

 

select count(t.t_id) from teacher t
where t.t_name like '李%'

 

#查詢學過"張三"老師授課的同窗的信息

select a.*
from student a
inner join score b on a.s_id = b.s_id
where b.c_id in(
    select c_id from course where t_id = (
        select t_id from teacher where t_name = '張三'
    )
)

 

#

#查詢沒學過"張三"老師授課的同窗的信息

select * from student c 
where c.s_id not in( 
    select a.s_id from student a 
    join score b on a.s_id=b.s_id 
    where b.c_id in( 
        select a.c_id from course a 
        join teacher b on a.t_id = b.t_id 
        where t_name ='張三'
    )
);

 

#查詢學過編號爲"01"而且也學過編號爲"02"的課程的同窗的信息

select a.* from student a
inner join score b on a.s_id = b.s_id
inner join score c on a.s_id = c.s_id
where b.c_id = '01'
and c.c_id = '02';

或者

select a.* from 
student a,score b,score c 
where a.s_id = b.s_id  
and a.s_id = c.s_id 
and b.c_id='01' 
and c.c_id='02';

 

#查詢學過編號爲"01"可是沒有學過編號爲"02"的課程的同窗的信息

select * from student 
where s_id in (select s_id from score where c_id='01')
and s_id not in (select s_id from score where c_id='02')

 

#查詢沒有學全全部課程的同窗的信息

select a.* from student a
left join score b on a.s_id = b.s_id
group by a.s_id
having count(b.s_id) < (select count(c_id) from course);

 

#查詢至少有一門課與學號爲"01"的同窗所學相同的同窗的信息

select a.* from student a
inner join score b on a.s_id = b.s_id
where b.c_id in (
    select c_id from score where s_id ='01'
)
group by a.s_id;

 

#查詢和"01"號的同窗學習的課程徹底相同的其餘同窗的信息

select a.* from student a
inner join score b on a.s_id = b.s_id
where a.s_id <> '01'
group by b.s_id
having group_concat(b.c_id ORDER BY c_id) = 
    (
        select group_concat(c_id ORDER BY c_id) 
        from score where s_id = '01'
    );

 

#查詢沒學過"張三"老師講授的任一門課程的學生姓名

select s_name from student where s_id not in (
    select distinct s_id from score where c_id in (
        select c_id from course where t_id = (select t_id from teacher where t_name ='張三')
    )
);

 

 #查詢兩門及其以上不及格課程的同窗的學號，姓名及其平均成績

select a.s_id,a.s_name,round(avg(b.s_score),2) as avg_score from student a
left join score b on a.s_id = b.s_id
where a.s_id in (
    select s_id from score
    where s_score < 60 
    group by s_id 
    having count(s_id) >=2
)
group by a.s_id,a.s_name;

 

#檢索"01"課程分數小於60，按分數降序排列的學生信息

select a.* from student a
inner join score b on a.s_id = b.s_id
where b.s_score < 60
and b.c_id = '01'
order by b.s_score desc;

 

#按平均成績從高到低顯示全部學生的全部課程的成績以及平均成績

select a.s_id,
    (select s_score from score where s_id=a.s_id and c_id='01') as 語文, 
    (select s_score from score where s_id=a.s_id and c_id='02') as 數學, 
    (select s_score from score where s_id=a.s_id and c_id='03') as 英語, 
    round(avg(s_score),2) as 平均分 
    from score a 
    GROUP BY a.s_id 
    ORDER BY 平均分 DESC;

 

 

 #查詢各科成績最高分、最低分和平均分：以以下形式顯示：課程ID，課程name，最高分，最低分，平均分，及格率，中等率，優良率，優秀率 --及格爲>=60，中等爲：70-80，優良爲：80-90，優秀爲：>=90

select a.c_name,max(b.s_score) as max_score,
    min(b.s_score) as min_score,
    round(avg(b.s_score),2) as avg_score ,
    round(100*sum(case when b.s_score >= 60 then 1 else 0 end)/sum(case when b.s_score then 1 else 0 end),2) as 及格率,
    round(100*sum(case when b.s_score >= 70 then 1 else 0 end)/sum(case when b.s_score then 1 else 0 end),2) as 中等率,
    round(100*sum(case when b.s_score >= 80 then 1 else 0 end)/sum(case when b.s_score then 1 else 0 end),2) as 優良率,
    round(100*sum(case when b.s_score >= 90 then 1 else 0 end)/sum(case when b.s_score then 1 else 0 end),2) as 優秀率
from course a
left join score b on a.c_id = b.c_id
group by a.c_id;

 

#查詢學生的總成績並進行排名

#

select * from (
    select a.s_id,sum(b.s_score) as total_score from student a
    left join score b on a.s_id = b.s_id
    group by a.s_id
)m
order by m.total_score desc

 

#查詢不一樣老師所教不一樣課程平均分從高到低顯示

select a.t_id,a.t_name,b.c_id,round(avg(c.s_score),2) as avg_score 
from teacher a
left join course b on a.t_id = b.t_id
left join score c on b.c_id = c.c_id
group by a.t_id,b.c_id,a.t_name
order by avg_score desc;

 

#查詢全部課程的成績第2名到第3名的學生信息及該課程成績

select d.*,c.排名,c.s_score,c.c_id 
from ( 
    select a.s_id,a.s_score,a.c_id,@i:=@i+1 as 排名 
    from score a,
    (select @i:=0)s where a.c_id='01' ORDER BY a.s_score DESC )c 
left join student d on c.s_id=d.s_id 
where 排名 BETWEEN 2 AND 3 
UNION 
select d.*,c.排名,c.s_score,c.c_id 
from ( 
    select a.s_id,a.s_score,a.c_id,@j:=@j+1 as 排名 from 
    score a,
    (select @j:=0)s where a.c_id='02' ORDER BY a.s_score DESC 
)c 
left join student d on c.s_id=d.s_id 
where 排名 BETWEEN 2 AND 3 
UNION 
select d.*,c.排名,c.s_score,c.c_id 
from ( 
    select a.s_id,a.s_score,a.c_id,@k:=@k+1 as 排名 
    from score a,
    (select @k:=0)s where a.c_id='03' ORDER BY a.s_score DESC 
)c 
left join student d on c.s_id=d.s_id 
where 排名 BETWEEN 2 AND 3;

 

#統計各科成績各分數段人數：課程編號,課程名稱,[100-85],[85-70],[70-60],[0-60]及所佔百分比

select a.c_id,a.c_name,
        sum(case when b.s_score > 85  and b.s_score <=100 then 1 else 0 end) as '85-100人數',
        round(100*sum(case when b.s_score > 85  and b.s_score <=100 then 1 else 0 end)/sum(case when b.s_score then 1 else 0 end),2) as '85-100比例',
        sum(case when b.s_score > 70  and b.s_score <=85 then 1 else 0 end) as '70-85人數',
    round(100*sum(case when b.s_score > 70  and b.s_score <=85 then 1 else 0 end)/sum(case when b.s_score then 1 else 0 end),2) as '70-85比例',
        sum(case when b.s_score > 60  and b.s_score <=70 then 1 else 0 end) as '60-70人數',
        round(100*sum(case when b.s_score > 60  and b.s_score <=70 then 1 else 0 end)/sum(case when b.s_score then 1 else 0 end),2) as '60-70比例',
        sum(case when b.s_score > 0  and b.s_score <=60 then 1 else 0 end) as '0-60人數',
    round(100*sum(case when b.s_score > 0  and b.s_score <=60 then 1 else 0 end)/sum(case when b.s_score then 1 else 0 end),2) as '0-60比例'
from course a
left join score b on a.c_id = b.c_id
group by a.c_id,a.c_name;

 

#查詢學平生均成績及其名次

select a.s_id,a.s_name,
        round(avg(b.s_score),2) as avg_score
from student a
left join score b on a.s_id = b.s_id
group by a.s_id,a.s_name
order by b.s_score desc ;

 

#查詢每門課程被選修的學生數

select c.c_id,count(s.s_id) from course c
left join score s on c.c_id = s.c_id
group by c.c_id;

 

#查詢出只有兩門課程的所有學生的學號和姓名

 

轉自：https://blog.csdn.net/fashion2014/article/details/78826299