xinmo_jin的博客：sql语句练习50题(Mysql版)

当前位置博文首页 > xinmo_jin的博客：sql语句练习50题(Mysql版)

xinmo_jin的博客：sql语句练习50题(Mysql版)

作者：[db:作者] 时间：2021-07-19 10:21

m0_50125842 2021-01-28 17:33:01 21 收藏
分类专栏： sql 文章标签： mysql ubuntu
版权

sql语句练习50题(Mysql版)

表名和字段

–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, 教师编号
–3.教师表
Teacher(t_id,t_name) --教师编号,教师姓名
–4.成绩表
Score(s_id,c_id,s_score) --学生编号,课程编号,分数
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)
);
–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)
);
–教师表
CREATE TABLE Teacher(
t_id VARCHAR(20),
t_name VARCHAR(20) NOT NULL DEFAULT ‘’,
PRIMARY KEY(t_id)
);
–成绩表
CREATE TABLE Score(
s_id VARCHAR(20),
c_id VARCHAR(20),
s_score INT(3),
PRIMARY KEY(s_id,c_id)
);
–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’);

–教师表测试数据
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);

练习题和sql语句

– 1、查询"01"课程比"02"课程成绩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

–也可以这样写
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
– 2、查询"01"课程比"02"课程成绩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、查询平均成绩大于等于60分的同学的学生编号和学生姓名和平均成绩
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 avg_score >=60;

– 4、查询平均成绩小于60分的同学的学生编号和学生姓名和平均成绩
– (包括有成绩的和无成绩的)

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 avg_score <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、查询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、查询"李"姓老师的数量
select count(t_id) from teacher where t_name like ‘李%’;

– 7、查询学过"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、查询没学过"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 a.c_id from course a join teacher b on a.t_id = b.t_id where t_name =‘张三’));
– 9、查询学过编号为"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、查询学过编号为"01"但是没有学过编号为"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、查询没有学全所有课程的同学的信息
–@wendiepei的写法
select s.* from student s
left join Score s1 on s1.s_id=s.s_id
group by s.s_id having count(s1.c_id)<(select count() from course)
–@k1051785839的写法
select *
from student
where s_id not in(
select s_id from score t1
group by s_id having count() =(select count(distinct c_id) from course))
– 12、查询至少有一门课与学号为"01"的同学所学相同的同学的信息

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、查询和"01"号的同学学习的课程完全相同的其他同学的信息
–@ouyang_1993的写法
SELECT
Student.*
FROM
Student
WHERE
s_id IN (SELECT s_id FROM Score GROUP BY s_id HAVING COUNT(s_id) = (
#下面的语句是找到’01’同学学习的课程数
SELECT COUNT(c_id) FROM Score WHERE s_id = ‘01’
)
)
AND s_id NOT IN (
#下面的语句是找到学过‘01’同学没学过的课程，有哪些同学。并排除他们
SELECT s_id FROM Score
WHERE c_id IN(
#下面的语句是找到‘01’同学没学过的课程
SELECT DISTINCT c_id FROM Score
WHERE c_id NOT IN (
#下面的语句是找出‘01’同学学习的课程
SELECT c_id FROM Score WHERE s_id = ‘01’
)
) GROUP BY s_id
) #下面的条件是排除01同学
AND s_id NOT IN (‘01’)
–@k1051785839的写法
SELECT
t3.*
FROM
(
SELECT
s_id,
group_concat(c_id ORDER BY c_id) group1
FROM
score
WHERE
s_id <> ‘01’
GROUP BY
s_id
) t1
INNER JOIN (
SELECT
group_concat(c_id ORDER BY c_id) group2
FROM
score
WHERE
s_id = ‘01’
GROUP BY
s_id
) t2 ON t1.group1 = t2.group2
INNER JOIN student t3 ON t1.s_id = t3.s_id

– 14、查询没学过"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’)));

– 15、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 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 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;
–@喝完这杯还有一箱的写法
SELECT a.s_id,MAX(CASE a.c_id WHEN ‘01’ THEN a.s_score END ) 语文,
MAX(CASE a.c_id WHEN ‘02’ THEN a.s_score END ) 数学,
MAX(CASE a.c_id WHEN ‘03’ THEN a.s_score END ) 英语,
avg(a.s_score),b.s_name FROM Score a JOIN Student b ON a.s_id=b.s_id GROUP BY a.s_id ORDER BY 5 DESC
– 18.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
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、按各科成绩进行排序，并显示排名
– 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 GROUP BY s_id,c_id,s_score ORDER BY s_score DESC
)a,(select @k:=0,@i:=0,@score:=0)s
–@k1051785839的写法
(select * from (select
t1.c_id,
t1.s_score,
(select count(distinct t2.s_score) from score t2 where t2.s_score>=t1.s_score and t2.c_id=‘01’) rank
FROM score t1 where t1.c_id=‘01’
order by t1.s_score desc) t1)
union
(select * from (select
t1.c_id,
t1.s_score,
(select count(distinct t2.s_score) from score t2 where t2.s_score>=t1.s_score and t2.c_id=‘02’) rank
FROM score t1 where t1.c_id=‘02’
order by t1.s_score desc) t2)
union
(select * from (select
t1.c_id,
t1.s_score,
(select count(distinct t2.s_score) from score t2 where t2.s_score>=t1.s_score and t2.c_id=‘03’) rank
FROM score t1 where t1.c_id=‘03’
order by t1.s_score desc) t3)
– 20、查询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、查询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、查询所有课程的成绩第2名到第3名的学生信息及该课程成绩

	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;

– 23、统计各科成绩各分数段人数：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

1
2
3
4
5
6
7
8
9
10
11
12
13
14

– 24、查询学生平均成绩及其名次

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 ORDER BY avg_s DESC)a,(select @avg_score:=0,@i:=0,@k:=0)b;

1
2
3
4
5

– 25、查询各科成绩前三名的记录
– 1.选出b表比a表成绩大的所有组
– 2.选出比当前id成绩大的小于三个的
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)??
ORDER BY a.c_id,a.s_score DESC

– 26、查询每门课程被选修的学生数

select c_id,count(s_id) from score a GROUP BY c_id

1

– 27、查询出只有两门课程的全部学生的学号和姓名
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、查询男生、女生人数
select s_sex,COUNT(s_sex) as 人数 from student GROUP BY s_sex

– 29、查询名字中含有"风"字的学生信息

select * from student where s_name like '%风%';

1

– 30、查询同名同性学生名单，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

1
2
3

– 31、查询1990年出生的学生名单

select s_name from student where s_birth like '1990%'

1

– 32、查询每门课程的平均成绩，结果按平均成绩降序排列，平均成绩相同时，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、查询平均成绩大于等于85的所有学生的学号、姓名和平均成绩

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

1
2

– 34、查询课程名称为"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 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

1
2

– 35、查询所有学生的课程及分数情况；

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、查询任何一门课程成绩在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、查询不及格的课程
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、查询课程编号为01且课程成绩在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、查询选修"张三"老师所授课程的学生中，成绩最高的学生信息及其成绩

-- 查询老师id	
select c_id from course c,teacher d where c.t_id=d.t_id and d.t_name='张三'
-- 查询最高分（可能有相同分数）
select MAX(s_score) from score where c_id='02'
-- 查询信息
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='02')

– 41、查询不同课程成绩相同的学生的学生编号、课程编号、学生成绩
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、查询每门功成绩最好的前两名
– 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_id,count(*) as total from score GROUP BY c_id HAVING total>5 ORDER BY total,c_id ASC

– 44、检索至少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、查询选修了全部课程的学生信息
select * from student where s_id in(
select s_id from score GROUP BY s_id HAVING count()=(select count() from course))

–46、查询各学生的年龄
– 按照出生日期来算，当前月日 < 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、查询本周过生日的学生
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、查询下周过生日的学生
select * from student where WEEK(DATE_FORMAT(NOW(),’%Y%m%d’))+1 =WEEK(s_birth)

– 49、查询本月过生日的学生

select * from student where MONTH(DATE_FORMAT(NOW(),’%Y%m%d’)) =MONTH(s_birth)

– 50、查询下月过生日的学生
select * from student where MONTH(DATE_FORMAT(NOW(),’%Y%m%d’))+1 =MONTH(s_birth)

MySQL等 SQL语句在线练习
大数据开发
5万+
SQL语句在线练习 http://sample.jimstone.com.cn/xsql/ SQL语句测验 http://www.w3school.com.cn/quiz/quiz.asp?quiz=sql sql语句练习50题(Mysql版) https://blog.csdn.net/fashion2014/article/details/78826299/ htt…
表情包

Zoro_16
Zoro_16:第一题差点给我劝退了…2 年前回复
[点赞] 87
qq_41795056
码工 [码工] 宫廷白开水回复bmync:

    SELECT
    	st.s_id,st.s_name,st.s_birth,st.s_sex,sc.s_score as score1,scc.s_score as score2
    FROM
    	Student st
    	LEFT JOIN Score sc ON st.s_id = sc.s_id AND sc.c_id = '01'
    	LEFT JOIN Score scc ON st.s_id = scc.s_id And scc.c_id = '02'
    	WHERE IFNULL(sc.s_score,0) > IFNULL(scc.s_score,0)

是01课比02课分数大，不过不伤大雅1 月前回复
[点赞]
bmync
bmync回复杰小生:

    SELECT
    	* 
    FROM
    	Student st
    	LEFT JOIN Score sc ON st.s_id = sc.s_id AND sc.c_id = '01'
    	LEFT JOIN Score scc ON st.s_id = scc.s_id And scc.c_id = '02'
    	WHERE IFNULL(sc.s_score,0) < IFNULL(scc.s_score,0)

两种思路，第一种查询时间太太太久了，还有感觉答案不够严谨，我用了一周的时间，太菜了1 月前回复
[点赞] 1
bmync
bmync回复杰小生:

    SELECT
    	* 
    FROM
    	(
    	SELECT
    		sts.*,
    		s.s_score 
    	FROM
    		( SELECT * FROM Student st CROSS JOIN Course c ) sts
    		LEFT JOIN Score s ON sts.s_id = s.s_id 
    		AND s.c_id = sts.c_id 
    	) AS datatable
    	JOIN (
    	SELECT
    		sts.*,
    		s.s_score 
    	FROM
    		( SELECT * FROM Student st CROSS JOIN Course c ) sts
    		LEFT JOIN Score s ON sts.s_id = s.s_id 
    		AND s.c_id = sts.c_id 
    	) AS aa ON aa.s_id = datatable.s_id 
    	AND datatable.c_id = '01' 
    	AND aa.c_id = "02" 
    	AND IFNULL( datatable.s_score, 0 ) > IFNULL(
    	aa.s_score,
    	0)

1 月前回复
[点赞]
weixin_43673660
杰小生回复杰小生:傻瓜我,这不就行了:

    select Student.s_id 学生编号,Student.s_name 学生姓名,A.`平均成绩` from 
    (SELECT s_id,avg(s_score) 平均成绩 FROM Score  group by s_id having AVG(s_score) >= 60) A 
    left join Student
    on A.s_id = Student.s_id;

2 月前回复
[点赞]
weixin_43673660
杰小生回复杰小生:

    -- 3、查询平均成绩大于等于60分的同学的学生编号和学生姓名和平均成绩
    select Student.s_id 学生编号,Student.s_name 学生姓名,A.`平均成绩` from 
    (SELECT Student.s_name,avg(s_score) 平均成绩 FROM Score LEFT JOIN Student ON Student.s_id = Score.s_id group by s_name having AVG(s_score) >= 60) A 
    left join Student
    on A.s_name = Student.s_name;

2 月前回复
[点赞]
weixin_43673660
杰小生回复杰小生:

    -- 2、查询"01"课程比"02"课程成绩低的学生的信息及课程分数
    -- 思路与第一题一致,条件换下就行了
    select A.s_name,语文,数学 from 
    (SELECT c_id,s_score 语文,s_name FROM Score LEFT JOIN Student ON Student.s_id = Score.s_id WHERE c_id = 01)  A
    join
    (SELECT c_id,s_score 数学,s_name FROM Score LEFT JOIN Student ON Student.s_id = Score.s_id WHERE c_id = 02)  B
    on
    A.s_name = B.s_name where 语文 < 数学;

2 月前回复
[点赞]
weixin_43673660
杰小生回复杰小生:

    select A.s_name,语文,数学 from 
    (SELECT c_id,s_score 语文,s_name FROM Score LEFT JOIN Student ON Student.s_id = Score.s_id WHERE c_id = 01)  A
    join
    (SELECT c_id,s_score 数学,s_name FROM Score LEFT JOIN Student ON Student.s_id = Score.s_id WHERE c_id = 02)  B
    on
    A.s_name = B.s_name where 语文 < 数学;

2 月前回复
[点赞]
weixin_43673660
杰小生回复:

    -- 1、查询"01"课程比"02"课程成绩高的学生的信息及课程分数
    select A.s_name,语文,数学 from 
    (SELECT c_id,s_score 语文,s_name FROM Score LEFT JOIN Student ON Student.s_id = Score.s_id WHERE c_id = 01)  A
    join
    (SELECT c_id,s_score 数学,s_name FROM Score LEFT JOIN Student ON Student.s_id = Score.s_id WHERE c_id = 02)  B
    on
    A.s_name = B.s_name where 语文 >数学;

2 月前回复
[点赞]
weixin_43673660
杰小生回复:已经劝退了2 月前回复
[点赞]
sinat_37676560
码哥 [码哥] chen_mg回复码上圆梦:哈哈哈，从后面做起，咸鱼翻身4 月前回复
[点赞]
weixin_44897262
码上圆梦回复窝窝头，:豪气风发来找题,第一题干废了...5 月前回复
[点赞] 2
weixin_44844568
几月的云回复:+16 月前回复
[点赞]
weixin_48992500
weixin_48992500回复:+16 月前回复
[点赞]
zhccjun
zhccjun回复:第一题竟然是我前几天的笔试题 ，上次没做出来这次也没做出来，嗨呀。7 月前回复
[点赞] 2
qq_42195733
那就试试就逝世回复:太TM真实了7 月前回复
[点赞] 1
qq_41093779
优秀的仙回复:就看了第一题我就知道我啥水平了8 月前回复
[点赞] 2
treblez
码哥 [码哥] treble-z回复车大炮鸡仔:学一下内联外联就好了9 月前回复
[点赞]
jinyuheng1987
车大炮鸡仔回复:第一题的第一个查询方法我还真看不懂，第二方方法还好。10 月前回复
[点赞] 1
valkyrja110
valkyrja110回复:第一题想了1个小时，才弄出来。。。。。。终于知道自己是什么水平了10 月前回复
[点赞]
weixin_45046849
weixin_45046849回复: +1才知道自连接可以一个表当俩个表用2 年前回复
[点赞]
weixin_39941448
好大的风啊回复: 卧槽，我以为只有我才这么认为2 年前回复
[点赞] 1
qq_34713259
俄亥俄州还给鹅回复: +12 年前回复
[点赞]
Adongbb
Adong董雪鹏回复: +100862 年前回复
[点赞]
qq_42538451
榴莲啊千层回复窝窝头，: +12 年前回复
[点赞] 1
Oh_Sun
窝窝头，回复: 同感 哈哈哈哈2 年前回复
[点赞] 1

L_Dragon66
L_Dragon66:第4题这样写简洁些吧

select
    S.s_id,
    S.s_name,
    round(avg(ifnull(C.s_score,0)),2) as avg_score
from
    Student S
left join Score C
on S.s_id = C.s_id
group by s_id
having avg_score < 60;

2 年前回复
[点赞] 14
qq_38859454
九歌_jason回复danny_hh:对6 月前回复
[点赞] 1
jiulou_1965
jiulou_1965回复danny_hh:它有ifnull7 月前回复
[点赞]
danny_hh
danny_hh回复:#4、查询平均成绩小于60分的同学的学生编号和学生姓名和平均成绩(包括有成绩的和无成绩的) 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 HAVING avg_score < 60 or avg_score is null; 添加个null的判断才全8 月前回复
[点赞] 1

Ohuo_Chen
二三哟:13题：

SELECT
 Student.*
FROM
 Student
WHERE
 s_id IN (select s_id from score GROUP BY s_id HAVING GROUP_CONCAT(c_id) = (    
     SELECT GROUP_CONCAT(c_id) FROM Score WHERE s_id = '01')
 );

2 年前回复
[点赞] 6
xwxyxxubo
Snails_Rocket回复:大佬，都没见过GROUP_CONCAT()函数2 月前回复
[点赞]
weixin_43992119
此湖不结冰回复:你这个需要对c_id 经行排序这样才能保证相等时是一样的7 月前回复
[点赞] 1
happenwhat
second_ming回复:这个结果没有排除学生编号为“01”，你可以再看下1 年前回复
[点赞]
three_thing
three_thing回复: 大佬，这里有个问题，就是你在聚集的时候，01 学号生成的是01，02，03，但是如果02学号的学生生成的可能是02，01，03，课程一样，但是结果不相等2 年前回复
[点赞] 2
three_thing
three_thing回复: 大佬，受我一拜2 年前回复
[点赞]
weixin_38688399
xxZkj回复: 大佬给跪了2 年前回复
[点赞]
Zero_Dream2019
Zero_Dream2019回复: 牛逼牛逼2 年前回复
[点赞]
aishuijiaodemao
练不出腹肌不改头像回复: 大佬受我一拜2 年前回复
[点赞]
qq_35297408
qq_35297408回复: 牛逼,学习了2 年前回复
[点赞]
User_Yr
全球变暖i回复: HAVING那一块儿改成 HAVING s_id <> ‘01’ AND GROUP_CONCAT(c_id)等等2 年前回复
[点赞] 1

three_thing
three_thing:13题: 参考：二三哟，他没有注意到如果c_id 顺序不一样的情况，比如 01号学生课程拼接顺序是 01，02,03，但是02号学生是02,01,03，用他的sql会出现问题，因此每次拼接都进行排序拼接，修改后代码如下：

SELECT
	*
FROM
	student
WHERE
	s_id IN (SELECT s_id FROM score GROUP BY s_id HAVING group_concat(c_id ORDER BY c_id) = (
				SELECT group_concat(c.c_id ORDER BY c_id) FROM score c WHERE c.s_id = '01' ));

2 年前回复
[点赞] 5
a1385341
水手怕水回复second_ming:默认数据不排序没问题，考虑现实情况得排序2 月前回复
[点赞]
qq_27817541
qq_27817541回复second_ming:您好，我这边测试下来不排序的话是有问题的…4 月前回复
[点赞]
happenwhat
second_ming回复:已经测试，在group_concat 内部不需要排序，二三呦的回答，除了没有过滤“01“，其他都ok1 年前回复

上一篇：昆山玉：2016年第07届蓝桥杯 Java B组决赛真题详解及小结

下一篇：没有了

立即下载 - IIS7 站长工具包