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SQL 练习
1、表构造
–1. 学生表
Student(s_id,s_name,s_birth,s_sex) –学生编号, 学生姓名, 出生年月, 学生性别
–2. 课程表
Course(c_id,c_name,t_id) – –课程编号, 课程名称, 老师编号
–3. 老师表
Teacher(t_id,t_name) –老师编号, 老师姓名
–4. 成绩表
Score(s_id,c_id,s_score) –学生编号, 课程编号, 分数 2、测试数据
-- 建表
-- 学生表
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`)
) ENGINE=InnoDB DEFAULT CHARSET=utf8;
-- 课程表
CREATE TABLE `Course`(`c_id` VARCHAR(20),
`c_name` VARCHAR(20) NOT NULL DEFAULT '',
`t_id` VARCHAR(20) NOT NULL,
PRIMARY KEY(`c_id`)
) ENGINE=InnoDB DEFAULT CHARSET=utf8;
-- 老师表
CREATE TABLE `Teacher`(`t_id` VARCHAR(20),
`t_name` VARCHAR(20) NOT NULL DEFAULT '',
PRIMARY KEY(`t_id`)
) ENGINE=InnoDB DEFAULT CHARSET=utf8;
-- 成绩表
CREATE TABLE `Score`(`s_id` VARCHAR(20),
`c_id` VARCHAR(20),
`s_score` INT(3),
PRIMARY KEY(`s_id`,`c_id`)
) ENGINE=InnoDB DEFAULT CHARSET=utf8;
-- 插入学生表测试数据
insert into Student values('01' , '赵雷' , '1990-01-01' , '男');
insert into Student values('02' , '钱电' , '1990-12-21' , '男');
insert into Student values('03' , '孙风' , '1990-05-20' , '男');
insert into Student values('04' , '李云' , '1990-08-06' , '男');
insert into Student values('05' , '周梅' , '1991-12-01' , '女');
insert into Student values('06' , '吴兰' , '1992-03-01' , '女');
insert into Student values('07' , '郑竹' , '1989-07-01' , '女');
insert into Student values('08' , '王菊' , '1990-01-20' , '女');
-- 课程表测试数据
insert into Course values('01' , '语文' , '02');
insert into Course values('02' , '数学' , '01');
insert into Course values('03' , '英语' , '03');
-- 老师表测试数据
insert into Teacher values('01' , '张三');
insert into Teacher values('02' , '李四');
insert into Teacher values('03' , '王五');
-- 成绩表测试数据
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);3、测试题
-- 1、查问 "01" 课程比 "02" 课程问题高的学生的信息及课程分数
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' or c.c_id = NULL where b.s_score>c.s_score
-- 2、查问 "01" 课程比 "02" 课程问题低的学生的信息及课程分数
select a.* ,b.s_score as 01_score,c.s_score as 02_score from
student a left join score b on a.s_id=b.s_id and b.c_id='01' or b.c_id=NULL
join score c on a.s_id=c.s_id and c.c_id='02' where b.s_score<c.s_score
-- 3、查问均匀问题大于等于 60 分的同学的学生编号和学生姓名和均匀问题
select b.s_id,b.s_name,ROUND(AVG(a.s_score),2) as avg_score from
student b
join score a on b.s_id = a.s_id
GROUP BY b.s_id,b.s_name HAVING ROUND(AVG(a.s_score),2)>=60;
-- 4、查问均匀问题小于 60 分的同学的学生编号和学生姓名和均匀问题
-- (包含有问题的和无问题的)
select b.s_id,b.s_name,ROUND(AVG(a.s_score),2) as avg_score from
student b
left join score a on b.s_id = a.s_id
GROUP BY b.s_id,b.s_name HAVING ROUND(AVG(a.s_score),2)<60
union
select a.s_id,a.s_name,0 as avg_score from
student a
where a.s_id not in (select distinct s_id from score);
-- 5、查问所有同学的学生编号、学生姓名、选课总数、所有课程的总成绩
select a.s_id,a.s_name,count(b.c_id) as sum_course,sum(b.s_score) as sum_score from
student a
left join score b on a.s_id=b.s_id
GROUP BY a.s_id,a.s_name;
-- 6、查问 "李" 姓老师的数量
select count(t_id) from teacher where t_name like '李 %';
-- 7、查问学过 "张三" 老师授课的同学的信息
select a.* from
student a
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 = '张三'));
-- 8、查问没学过 "张三" 老师授课的同学的信息
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 c_id from course where t_id =(select t_id from teacher where t_name = '张三')));
-- 9、查问学过编号为 "01" 并且也学过编号为 "02" 的课程的同学的信息
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';
-- 10、查问学过编号为 "01" 然而没有学过编号为 "02" 的课程的同学的信息
select a.* from
student a
where a.s_id in (select s_id from score where c_id='01') and a.s_id not in(select s_id from score where c_id='02')
-- 11、查问没有学全所有课程的同学的信息
select s.* from
student s where s.s_id in(
select s_id from score where s_id not in(
select a.s_id from score a
join score b on a.s_id = b.s_id and b.c_id='02'
join score c on a.s_id = c.s_id and c.c_id='03'
where a.c_id='01'))
-- 12、查问至多有一门课与学号为 "01" 的同学所学雷同的同学的信息
select * from student where s_id in(select distinct a.s_id from score a where a.c_id in(select a.c_id from score a where a.s_id='01')
);
-- 13、查问和 "01" 号的同学学习的课程完全相同的其他同学的信息
select a.* from student a where a.s_id in(select distinct s_id from score where s_id!='01' and c_id in(select c_id from score where s_id='01')
group by s_id
having count(1)=(select count(1) from score where s_id='01'));
-- 14、查问没学过 "张三" 老师讲授的任一门课程的学生姓名
select a.s_name from student a where a.s_id not in (
select s_id from score where c_id =
(select c_id from course where t_id =(select t_id from teacher where t_name = '张三'))
group by s_id);
-- 15、查问两门及其以上不及格课程的同学的学号，姓名及其均匀问题
select a.s_id,a.s_name,ROUND(AVG(b.s_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(1)>=2)
GROUP BY a.s_id,a.s_name
-- 16、检索 "01" 课程分数小于 60，按分数降序排列的学生信息
select a.*,b.c_id,b.s_score from
student a,score b
where a.s_id = b.s_id and b.c_id='01' and b.s_score<60 ORDER BY b.s_score DESC;
-- 17、按均匀问题从高到低显示所有学生的所有课程的问题以及均匀问题
select a.s_id,(select s_score from score where s_id=a.s_id and c_id='01') as 语文,
(select s_score from score where s_id=a.s_id and c_id='02') as 数学,
(select s_score from score where s_id=a.s_id and c_id='03') as 英语,
round(avg(s_score),2) as 平均分 from score a GROUP BY a.s_id ORDER BY 平均分 DESC;
-- 18. 查问各科问题最高分、最低分和平均分：以如下模式显示：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，低劣为：80-90，优良为：>=90
select a.c_id,b.c_name,MAX(s_score),MIN(s_score),ROUND(AVG(s_score),2),
ROUND(100*(SUM(case when a.s_score>=60 then 1 else 0 end)/SUM(case when a.s_score then 1 else 0 end)),2) as 及格率,
ROUND(100*(SUM(case when a.s_score>=70 and a.s_score<=80 then 1 else 0 end)/SUM(case when a.s_score then 1 else 0 end)),2) as 中等率,
ROUND(100*(SUM(case when a.s_score>=80 and a.s_score<=90 then 1 else 0 end)/SUM(case when a.s_score then 1 else 0 end)),2) as 优良率,
ROUND(100*(SUM(case when a.s_score>=90 then 1 else 0 end)/SUM(case when a.s_score then 1 else 0 end)),2) as 优秀率
from score a left join course b on a.c_id = b.c_id GROUP BY a.c_id,b.c_name
-- 19、按各科问题进行排序，并显示排名 (实现不齐全)
-- mysql 没有 rank 函数
select a.s_id,a.c_id,
@i:=@i +1 as i 保留排名,
@k:=(case when @score=a.s_score then @k else @i end) as rank 不保留排名,
@score:=a.s_score as score
from (select s_id,c_id,s_score from score WHERE c_id='01' GROUP BY s_id,c_id,s_score ORDER BY s_score DESC)a,(select @k:=0,@i:=0,@score:=0)s
union
select a.s_id,a.c_id,
@i:=@i +1 as i,
@k:=(case when @score=a.s_score then @k else @i end) as rank,
@score:=a.s_score as score
from (select s_id,c_id,s_score from score WHERE c_id='02' GROUP BY s_id,c_id,s_score ORDER BY s_score DESC)a,(select @k:=0,@i:=0,@score:=0)s
union
select a.s_id,a.c_id,
@i:=@i +1 as i,
@k:=(case when @score=a.s_score then @k else @i end) as rank,
@score:=a.s_score as score
from (select s_id,c_id,s_score from score WHERE c_id='03' GROUP BY s_id,c_id,s_score ORDER BY s_score DESC)a,(select @k:=0,@i:=0,@score:=0)s
-- 20、查问学生的总成绩并进行排名
select a.s_id,
@i:=@i+1 as i,
@k:=(case when @score=a.sum_score then @k else @i end) as rank,
@score:=a.sum_score as score
from (select s_id,SUM(s_score) as sum_score from score GROUP BY s_id ORDER BY sum_score DESC)a,
(select @k:=0,@i:=0,@score:=0)s
-- 21、查问不同老师所教不同课程平均分从高到低显示
select a.t_id,c.t_name,a.c_id,ROUND(avg(s_score),2) as avg_score from course a
left join score b on a.c_id=b.c_id
left join teacher c on a.t_id=c.t_id
GROUP BY a.c_id,a.t_id,c.t_name ORDER BY avg_score DESC;
-- 22、查问所有课程的问题第 2 名到第 3 名的学生信息及该课程问题
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'
)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'
)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'
)c
left join student d on c.s_id=d.s_id
where 排名 BETWEEN 2 AND 3;
-- 23、统计各科问题各分数段人数：课程编号, 课程名称,[100-85],[85-70],[70-60],[0-60] 及所占百分比
select distinct f.c_name,a.c_id,b.`85-100`,b. 百分比,c.`70-85`,c. 百分比,d.`60-70`,d. 百分比,e.`0-60`,e. 百分比 from score a
left join (select c_id,SUM(case when s_score >85 and s_score <=100 then 1 else 0 end) as `85-100`,
ROUND(100*(SUM(case when s_score >85 and s_score <=100 then 1 else 0 end)/count(*)),2) as 百分比
from score GROUP BY c_id)b on a.c_id=b.c_id
left join (select c_id,SUM(case when s_score >70 and s_score <=85 then 1 else 0 end) as `70-85`,
ROUND(100*(SUM(case when s_score >70 and s_score <=85 then 1 else 0 end)/count(*)),2) as 百分比
from score GROUP BY c_id)c on a.c_id=c.c_id
left join (select c_id,SUM(case when s_score >60 and s_score <=70 then 1 else 0 end) as `60-70`,
ROUND(100*(SUM(case when s_score >60 and s_score <=70 then 1 else 0 end)/count(*)),2) as 百分比
from score GROUP BY c_id)d on a.c_id=d.c_id
left join (select c_id,SUM(case when s_score >=0 and s_score <=60 then 1 else 0 end) as `0-60`,
ROUND(100*(SUM(case when s_score >=0 and s_score <=60 then 1 else 0 end)/count(*)),2) as 百分比
from score GROUP BY c_id)e on a.c_id=e.c_id
left join course f on a.c_id = f.c_id
-- 24、查问学生均匀问题及其名次
select a.s_id,
@i:=@i+1 as '不保留空缺排名',
@k:=(case when @avg_score=a.avg_s then @k else @i end) as '保留空缺排名',
@avg_score:=avg_s as '平均分'
from (select s_id,ROUND(AVG(s_score),2) as avg_s from score GROUP BY s_id)a,(select @avg_score:=0,@i:=0,@k:=0)b;
-- 25、查问各科问题前三名的记录
-- 1. 选出 b 表比 a 表问题大的所有组
-- 2. 选出比以后 id 问题大的 小于三个的
select a.s_id,a.c_id,a.s_score from score a
left join score b on a.c_id = b.c_id and a.s_score<b.s_score
group by a.s_id,a.c_id,a.s_score HAVING COUNT(b.s_id)<3
ORDER BY a.c_id,a.s_score DESC
-- 26、查问每门课程被选修的学生数
select c_id,count(s_id) from score a GROUP BY c_id
-- 27、查问出只有两门课程的全副学生的学号和姓名
select s_id,s_name from student where s_id in(select s_id from score GROUP BY s_id HAVING COUNT(c_id)=2);
-- 28、查问男生、女生人数
select s_sex,COUNT(s_sex) as 人数 from student GROUP BY s_sex
-- 29、查问名字中含有 "风" 字的学生信息
select * from student where s_name like '% 风 %';
-- 30、查问同名同性学生名单，并统计同名人数
select a.s_name,a.s_sex,count(*) from student a JOIN
student b on a.s_id !=b.s_id and a.s_name = b.s_name and a.s_sex = b.s_sex
GROUP BY a.s_name,a.s_sex
-- 31、查问 1990 年出世的学生名单
select s_name from student where s_birth like '1990%'
-- 32、查问每门课程的均匀问题，后果按均匀问题降序排列，均匀问题雷同时，按课程编号升序排列
select c_id,ROUND(AVG(s_score),2) as avg_score from score GROUP BY c_id ORDER BY avg_score DESC,c_id ASC
-- 33、查问均匀问题大于等于 85 的所有学生的学号、姓名和均匀问题
select a.s_id,b.s_name,ROUND(avg(a.s_score),2) as avg_score from score a
left join student b on a.s_id=b.s_id GROUP BY s_id HAVING avg_score>=85
-- 34、查问课程名称为 "数学"，且分数低于 60 的学生姓名和分数
select a.s_name,b.s_score from score b LEFT JOIN student a on a.s_id=b.s_id where b.c_id=(select c_id from course where c_name ='数学') and b.s_score<60
-- 35、查问所有学生的课程及分数状况；select a.s_id,a.s_name,
SUM(case c.c_name when '语文' then b.s_score else 0 end) as '语文',
SUM(case c.c_name when '数学' then b.s_score else 0 end) as '数学',
SUM(case c.c_name when '英语' then b.s_score else 0 end) as '英语',
SUM(b.s_score) as '总分'
from student a left join score b on a.s_id = b.s_id
left join course c on b.c_id = c.c_id
GROUP BY a.s_id,a.s_name
-- 36、查问任何一门课程问题在 70 分以上的姓名、课程名称和分数；select a.s_name,b.c_name,c.s_score from course b left join score c on b.c_id = c.c_id
left join student a on a.s_id=c.s_id where c.s_score>=70
-- 37、查问不及格的课程
select a.s_id,a.c_id,b.c_name,a.s_score from score a left join course b on a.c_id = b.c_id
where a.s_score<60
-- 38、查问课程编号为 01 且课程问题在 80 分以上的学生的学号和姓名；select a.s_id,b.s_name from score a LEFT JOIN student b on a.s_id = b.s_id
where a.c_id = '01' and a.s_score>80
-- 39、求每门课程的学生人数
select count(*) from score GROUP BY c_id;
-- 40、查问选修 "张三" 老师所授课程的学生中，问题最高的学生信息及其问题
-- 查问老师 id
select c_id from course c,teacher d where c.t_id=d.t_id and d.t_name='张三'
-- 查问最高分（可能有雷同分数）select MAX(s_score) from score where c_id='02'
-- 查问信息
select a.*,b.s_score,b.c_id,c.c_name from Student a
LEFT JOIN Score b on a.s_id = b.s_id
LEFT JOIN Course c on b.c_id=c.c_id
where b.c_id =(select c_id from Course c,Teacher d where c.t_id=d.t_id and d.t_name='张三')
and b.s_score in (select MAX(s_score) from Score where c_id=(select c_id from Course c,Teacher d where c.t_id=d.t_id and d.t_name='张三'))
-- 41、查问不同课程问题雷同的学生的学生编号、课程编号、学生问题
select DISTINCT b.s_id,b.c_id,b.s_score from score a,score b where a.c_id != b.c_id and a.s_score = b.s_score
-- 42、查问每门功问题最好的前两名
-- 牛逼的写法
select a.s_id,a.c_id,a.s_score from Score a
where (select COUNT(1) from Score b where b.c_id=a.c_id and b.s_score>=a.s_score)<=2 ORDER BY a.c_id
-- 43、统计每门课程的学生选修人数（超过 5 人的课程才统计）。要求输入课程号和选修人数，查问后果按人数降序排列，若人数雷同，按课程号升序排列
select c_id,count(*) as total from score GROUP BY c_id HAVING total>5 ORDER BY total,c_id ASC
-- 44、检索至多选修两门课程的学生学号
select s_id,count(*) as sel from score GROUP BY s_id HAVING sel>=2
-- 45、查问选修了全副课程的学生信息
select * from student where s_id in(select s_id from score GROUP BY s_id HAVING count(*)=(select count(*) from course))
-- 46、查问各学生的年龄
-- 依照出生日期来算，以后月日 < 出生年月的月日则，年龄减一
select s_birth,(DATE_FORMAT(NOW(),'%Y')-DATE_FORMAT(s_birth,'%Y') -
(case when DATE_FORMAT(NOW(),'%m%d')>DATE_FORMAT(s_birth,'%m%d') then 0 else 1 end)) as age
from student;
-- 47、查问本周过生日的学生
select * from student where WEEK(DATE_FORMAT(NOW(),'%Y%m%d'))=WEEK(s_birth)
select * from student where YEARWEEK(s_birth)=YEARWEEK(DATE_FORMAT(NOW(),'%Y%m%d'))
select WEEK(DATE_FORMAT(NOW(),'%Y%m%d'))
-- 48、查问下周过生日的学生
select * from student where WEEK(DATE_FORMAT(NOW(),'%Y%m%d'))+1 =WEEK(s_birth)
-- 49、查问本月过生日的学生
select * from student where MONTH(DATE_FORMAT(NOW(),'%Y%m%d')) =MONTH(s_birth)
-- 50、查问下月过生日的学生
select * from student where MONTH(DATE_FORMAT(NOW(),'%Y%m%d'))+1 =MONTH(s_birth)练习题问题
— 为什么用 left join 而且加判断 OR c.c_id = NULL? 右表数据为空为什么思考？
总结
思路
- 表连贯就是先 join 失去一个表，而后再 join 条件失去另一个表
- 个别应用 left join，过滤用 join
- 用 left join 时，关联条件须要判断右表数据有没有为 NULL 的
正文完
