目录
一、准备工作（建表、插入数据）：
二、SQL练习（50道）
1、查询"01"课程比"02"课程成绩高的学生的信息及课程分数
2、查询"01"课程比"02"课程成绩低的学生的信息及课程分数
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
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)
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
6、查询"李"姓老师的数量
7、询学过"张三"老师授课的同学的信息
8、查询没学过"张三"老师授课的同学的信息
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
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
11、查询没有学全所有课程的同学的信息
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
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
14、查询没学过"张三"老师讲授的任一门课程的学生姓名
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
16、检索"01"课程分数小于60，按分数降序排列的学生信息
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
18、查询各科成绩最高分、最低分、平均分、及格率（60分以上）、中等率（70-80）、优良率（80-90）、优秀率（90以上）
19、按各科成绩进行排序，并显示排名
20、查询学生的总成绩并进行排名
21、查询不同老师所教不同课程平均分从高到低显示
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
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
24、查询学生平均成绩及其名次
25、查询各科成绩前三名的记录
26、查询每门课程被选修的学生数
27、查询出只有两门课程的全部学生的学号和姓名
28、查询男生、女生人数
29、查询名字中含有"风"字的学生信息
30、查询同名同性学生名单，并统计同名人数
31、查询1990年出生的学生名单
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
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
34、查询课程名称为"数学"，且分数低于60的学生姓名和分数
35、查询所有学生的课程及分数情况
36、查询任何一门课程成绩在70分以上的学生姓名、课程名称和分数
37、查询课程不及格的学生
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
39、求每门课程的学生人数
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
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
42、查询每门课程成绩最好的前三名
43、统计每门课程的学生选修人数（超过5人的课程才统计）
44、检索至少选修两门课程的学生学号
45、查询选修了全部课程的学生信息
46、查询各学生的年龄(周岁)
47、查询本周过生日的学生
48、查询下周过生日的学生
49、查询本月过生日的学生
50、查询12月份过生日的学生
一、准备工作（建表、插入数据）：
drop table if exists student;
create table student(
s_id int,
s_name varchar(8),
s_birth date,
s_sex char(4)
);
insert into student values
(1,'赵雷','1990-01-01','男'),
(2,'钱电','1990-12-21','男'),
(3,'孙风','1990-05-20','男'),
(4,'李云','1990-08-06','男'),
(5,'周梅','1991-12-01','女'),
(6,'吴兰','1992-03-01','女'),
(7,'郑竹','1989-07-01','女'),
(8,'王菊','1990-01-20','女');
drop table if exists score;
create table score(
s_id int,
c_id int,
s_score int
);
insert into score values
(1,1,80),
(1,2,90),
(1,3,99),
(2,1,70),
(2,2,60),
(2,3,65),
(3,1,80),
(3,2,80),
(3,3,80),
(4,1,50),
(4,2,30),
(4,3,40),
(5,1,76),
(5,2,87),
(6,1,31),
(6,3,34),
(7,2,89),
(7,3,98);
drop table if exists course;
create table course(
c_id int,
c_name varchar(8),
t_id int
);
insert into course values
(1,'语文',2),
(2,'数学',1),
(3,'英语',3);
drop table if exists teacher;
create table teacher(
t_id int,
t_name varchar(8)
);
insert into teacher values
(1,'张三'),
(2,'李四'),
(3,'王五');二、SQL练习（50道）
1、查询"01"课程比"02"课程成绩高的学生的信息及课程分数
-- 方法1
select s.*, sc1.s_score as score_01, sc2.s_score as score_02
from student s,
(select s_id, s_score from score where c_id = 1) sc1,
(select s_id, s_score from score where c_id = 2) sc2
where sc1.s_score > sc2.s_score
and sc1.s_id = sc2.s_id
and sc1.s_id = s.s_id
-- 方法2
select s.*, sc1.s_score score_01, sc2.s_score score_02
from score sc1, score sc2, student s
where sc1.s_id = sc2.s_id
and sc1.c_id = 1
and sc2.c_id = 2
and sc1.s_score > sc2.s_score
and s.s_id = sc1.s_id;
-- 方法3
select s.*,sc1.s_score score_01,sc2.s_score score02
from student s
inner join (select * from score where c_id = 1) sc1
on s.s_id = sc1.s_id
inner join (select * from score where c_id = 2) sc2
on s.s_id = sc2.s_id
where sc1.s_score > sc2.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
同(1)
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
-- 方法1
select s.s_id, s_name, round(avg(s_score),2) avg_score
from student s, score
where s.s_id = score.s_id
group by s.s_id,s_name
having avg_score >= 60;
-- 方法2
select s.s_id, s_name, avg_score
from student s,
(select s_id, round(avg(s_score),2) avg_score
from score
group by s_id
having avg_score >= 60) as temp
where s.s_id = temp.s_id;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)
-- 方法1 （先找到平均成绩小于60的学生信息，在用union连接，加上没有选课的学生信息）
select s.s_id, s_name, avg_score
from student s,
(select s_id, round(avg(s_score),2) avg_score
from score
group by s_id
having avg_score < 60) temp
where s.s_id = temp.s_id
union
select s_id, s_name, 0
from student
where s_id not in (select s_id from score)
-- 方法2 没有成绩即该学生没有选课， 查询结果需要包含成绩为空的学生信息， 使用外连接
select s.s_id, s_name,
(case when avg_score is null then 0 else avg_score end) avg_score
from student s
left join
(select s_id ,round(avg(s_score),2) avg_score
from score
group by s_id) temp
on temp.s_id = s.s_id
where avg_score < 60 or avg_score is null5、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
-- 方法1
select s.s_id, s_name, count(c_id) cnt_course,
(case when sum(s_score) is null then 0 else sum(s_score) end) sum_score
from student s
left join score
on score.s_id = s.s_id
group by s.s_id, s_name
-- 方法2
select s.s_id, s_name,
(case when cnt_course is null then 0 else cnt_course end) cnt_course,
(case when sum_score is null then 0 else sum_score end) sum_score
from student s
left join
(select s_id, count(c_id) cnt_course, sum(s_score) sum_score
from score
group by s_id) temp
on temp.s_id = s.s_id6、查询"李"姓老师的数量
select count(t_name) from teacher where t_name like '%李%';
7、询学过"张三"老师授课的同学的信息
-- 方法1（嵌套查询）
select *
from student
where s_id IN(
select s_id
from score
where c_id IN(
select c_id
from course
where t_id IN(
select t_id
from teacher
where t_name = '张三')));
-- 方法2（自然连接）
select student.*
from student natural join score
natural join course
natural join 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
where s_id NOT IN (
select s_id from score
where c_id IN(
select c_id from course
where t_id IN(
select t_id from teacher
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
-- 方法1
select * from student
where s_id IN (select s_id from score where c_id = 1)
and s_id IN (select s_id from score where c_id = 2)
-- 方法2
select s.*
from student s,
(select s_id from score where c_id = 1) sc1,
(select s_id from score where c_id = 2) sc2
where s.s_id = sc1.s_id
and s.s_id = sc2.s_id10、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
-- 方法1
select * from student
where s_id IN (select s_id from score where c_id = 1)
and s_id NOT IN (select s_id from score where c_id = 2)
-- 方法2（使用except，做减法, 包含条件1但不包含条件2的学生）
select s.*
from student s natural join score where c_id = 1
except
select s.*
from student s natural join score where c_id = 211、查询没有学全所有课程的同学的信息
-- 方法1 （用union连接 没选修完课程的学生id 和 没有选修课程的学生id）
select s.* from student s
where s_id IN (
select s_id
from score
group by s_id
having count(*) < (select count(*) from course)
)
union
select s.* from student s
where s_id NOT IN (
select s_id
from score
)
-- 方法2 查询选修全部课程的学生—————即不存在这样一个学生，该生没有
select * from student
where s_id IN (
select s_id from student s1
where NOT EXISTS (
select * from course
where NOT EXISTS (
select * from score
where score.s_id = s1.s_id
and score.c_id = course.c_id)))
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
-- 方法1
select * from student s
where s_id <> 1
and s_id IN (
select s_id from score
where c_id IN(
select c_id from score
where s_id = 1
)
)
-- 方法2
select distinct s.*
from student s, score sc1, score sc2
where s.s_id = sc1.s_id
and sc1.c_id = sc2.c_id
and sc2.s_id = 1
and sc1.s_id <> 113、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
-- (即 没有一门课程是01同学选了，而其他同学没有选的)
select s.* from student s
where s.s_id <> 1
and NOT EXISTS (
select * from score sc1
where sc1.s_id = 1
and NOT EXISTS (
select * from score sc2
where sc2.s_id = s.s_id
and sc2.c_id = sc1.c_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 * from student
where s_id NOT IN (
select s_id from score
where c_id IN(
select c_id from course
where t_id IN(
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
-- 方法1
select s.s_id, s_name, avg_score
from student s join(
select s_id , round(avg(s_score),2) avg_score
from score
where s_score < 60
group by s_id
having count(*) >= 2
)temp
on temp.s_id = s.s_id
)
-- 方法2
select s.s_id, s_name, round(avg(s_score),2) avg_score
from student s, score sc
where s.s_id = sc.s_id
and s_score < 60
group by s.s_id, s_name
having count(*) >= 2;16、检索"01"课程分数小于60，按分数降序排列的学生信息
-- 方法1
select s.*
from student s join (
select s_id, s_score
from score
where c_id = 1
and s_score < 60
)temp on temp.s_id = s.s_id
and s_score < 60
order by s_score desc
-- 方法2
select s.*
from student s
join score sc on sc.s_id = s.s_id
where c_id = 1
and s_score < 60
order by 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 s.s_id, s_name,
sum(case c_id when 1 then s_score else 0 end) 语文,
sum(case c_id when 2 then s_score else 0 end) 数学,
sum(case c_id when 3 then s_score else 0 end) 英语,
ifnull(round(avg(s_score),2),0) 平均成绩
from student s
left join score sc
on sc.s_id = s.s_id
group by s.s_id,s_name
order by 平均成绩 desc;
18、查询各科成绩最高分、最低分、平均分、及格率（60分以上）、中等率（70-80）、优良率（80-90）、优秀率（90以上）
select c.c_id 课程ID, c_name 课程名称,
max(s_score) 最高分, min(s_score) 最低分, round(avg(s_score),2) 平均分,
concat(round(sum(case when s_score >=60 then 1 else 0 end)/count(*) * 100, 2), '%') 及格率,
concat(round(sum(case when s_score between 70 and 80 then 1 else 0 end)/count(*) * 100, 2), '%') 中等率,
concat(round(sum(case when s_score between 80 and 90 then 1 else 0 end)/count(*) * 100, 2), '%') 优良率,
concat(round(sum(case when s_score >= 90 then 1 else 0 end)/count(*) * 100, 2), '%') 优秀率
from course c join score sc on sc.c_id = c.c_id
group by c.c_id, c_name19、按各科成绩进行排序，并显示排名
-- 先查询各科的排名，再将查询结果命名为temp, 和student,course连接
select c.c_id, c_name, s.s_id, s_name, s_score, srank
from student s, course c,
(select s_id, c_id, s_score,
1+(select count(*)
from score sc1
where sc1.c_id = sc2.c_id
and sc1.s_score > sc2.s_score
) srank
from score sc2
)temp
where temp.s_id = s.s_id
and temp.c_id = c.c_id
order by temp.c_id, s_score desc20、查询学生的总成绩并进行排名
select s.s_id, s_name, sum_score,
1+(select count(*)
from
(select s_id, sum(s_score) sum_score from score group by s_id) t1
where t1.sum_score > t2.sum_score
) srank
from (select s_id, sum(s_score) sum_score
from score
group by s_id) t2,
student s
where s.s_id = t2.s_id
order by srank asc
-- 方法2
WITH t as
(select s_id, sum(s_score) sum_score
from score
group by s_id
)
select s_name, t2.s_id, sum_score,
1+(select count(*)
from t t1
where t1.sum_score > t2.sum_score
) srank
from t t2, student s
where t2.s_id = s.s_id
order by srank
21、查询不同老师所教不同课程平均分从高到低显示
-- 方法1
select t.t_id, t_name, c.c_id, c_name, round(avg(s_score),2) avg_score
from score sc
join course c on c.c_id = sc.c_id
join teacher t on t.t_id = c.t_id
group by t.t_id, t_name, c.c_id, c_name
order by t.t_id, avg_score desc
-- 方法2
select t.t_id, t_name, c.c_id, c_name, avg_score
from (
select c_id, round(avg(s_score),2) avg_score
from score
group by c_id
) sc
join course c on c.c_id = sc.c_id
join teacher t on t.t_id = c.t_id
order by t.t_id, 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
-- 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 c.c_id, c_name, s.s_id, s_name, s_score, srank
from student s, course c,
(select s_id, c_id, s_score,
1+(select count(*)
from score sc1
where sc1.s_score > sc2.s_score
)srank
from score sc2
) temp
where s.s_id = temp.s_id
and c.c_id = temp.c_id
and srank in (2,3)
order by temp.c_id, s_score desc23、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 c.c_id 课程ID, c_name 课程名称,
concat(round(sum(case when s_score between 0 and 60 then 1 else 0 end)/count(*) * 100, 2), '%') 及格率,
sum(case when s_score between 0 and 60 then 1 else 0 end) 及格人数,
concat(round(sum(case when s_score between 60 and 70 then 1 else 0 end)/count(*) * 100, 2), '%') 中等率,
sum(case when s_score between 60 and 70 then 1 else 0 end) 中等人数,
concat(round(sum(case when s_score between 70 and 85 then 1 else 0 end)/count(*) * 100, 2), '%') 优良率,
sum(case when s_score between 70 and 85 then 1 else 0 end) 优良人数,
concat(round(sum(case when s_score between 85 and 100 then 1 else 0 end)/count(*) * 100, 2), '%') 优秀率,
sum(case when s_score between 85 and 100 then 1 else 0 end) 优秀人数
from course c join score sc on sc.c_id = c.c_id
group by c.c_id, c_name24、查询学生平均成绩及其名次
-- 先计算每个学生的平均成绩， 再用排名的sql语句
WITH t as(
select s_id, round(avg(s_score),2) avg_score
from score
group by s_id
)
select t2.s_id, s_name, avg_score,
1+ (
select count(*)
from t t1
where t1.avg_score> t2.avg_score
) srank
from t t2 natural join student
order by srank asc
25、查询各科成绩前三名的记录
select c_id, c_name, s_id, s_name, s_score
from student natural join score
natural join course
natural join (
select s_id, c_id, s_score
from score s1
where (
select count(*)
from score s2
where s2.c_id = s1.c_id
and s2.s_score > s1.s_score
) < 3
) temp
order by c_id, s_score desc26、查询每门课程被选修的学生数
-- 方法1
select c_id, c_name, count(*) cnt_studnet
from student s natural join score sc
natural join course c
group by c_id, c_name
-- 方法2
select c.c_id, c_name, cnt_student
from course c join (
select c_id, count(*) cnt_student
from score
group by c_id
) temp
on temp.c_id = c.c_id
27、查询出只有两门课程的全部学生的学号和姓名
-- 方法1
select s_id, s_name
from student
where s_id IN (
select s_id
from score
group by s_id
having count(*)=2
)
-- 方法2
select s_id, s_name
from student natural join score
group by s_id, s_name
having count(*) = 2
28、查询男生、女生人数
-- 方法1
select s_sex, count(*) cnt_sex
from student
group by s_sex;
-- 方法2
select sum(case s_sex when '男' then 1 else 0 end) 男生人数,
sum(case s_sex when '女' then 1 else 0 end) 女生人数
from student
29、查询名字中含有"风"字的学生信息
select * from student
where s_name like '%风%'
30、查询同名同性学生名单，并统计同名人数
select s_name, s_sex, count(*) cnt_student
from student
group by s_name, s_sex
having cnt_student > 1
31、查询1990年出生的学生名单
select * from student
where year(s_birth) = '1990'32、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
-- 方法1
select c_id, c_name, round(avg(s_score) ,2) avg_score
from course natural join score
group by c_id, c_name
order by avg_score desc, c_id asc
-- 方法2
select c.c_id, c_name, round(avg(s_score) ,2) avg_score
from course c
join score sc on sc.c_id = c.c_id
group by c.c_id, c_name
order by avg_score desc, c_id asc
-- 方法3
select c.c_id, c_name, avg_score
from course c join (
select c_id, round(avg(s_score) ,2) avg_score
from score
group by c_id
) temp
on temp.c_id = c.c_id
order by avg_score desc, c_id asc33、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
-- 方法1
select s_id, s_name, round(avg(s_score) ,2) avg_score
from student natural join score
group by s_id, s_name
having avg_score >= 85;
-- 方法2
select s.s_id, s_name, avg_score
from student s join (
select s_id, round(avg(s_score) ,2) avg_score
from score
group by s_id
) temp
on temp.s_id = s.s_id
where avg_score >= 85 34、查询课程名称为"数学"，且分数低于60的学生姓名和分数
-- 方法1
select s_name, s_score
from student natural join score
natural join course
where c_name = '数学'
and s_score < 60
-- 方法2
select s_name, s_score
from student s join (
select s_id, c_id, s_score
from score
where s_score < 60
) temp
on temp.s_id = s.s_id
join course c on c.c_id = temp.c_id
where c_name = '数学'35、查询所有学生的课程及分数情况
select s.s_id, s_name,
sum(case c_id when 1 then s_score else 0 end) 语文,
sum(case c_id when 2 then s_score else 0 end) 数学,
sum(case c_id when 3 then s_score else 0 end) 英语
from student s natural join score
group by s.s_id, s_name
36、查询任何一门课程成绩在70分以上的学生姓名、课程名称和分数
-- 任何一门成绩都要大于70， 则表示最低成绩也要大于70
select s_name, c_name, s_score
from student s natural join score
natural join course
where s_id IN (
select s_id
from score
group by s_id
having min(s_score) > 70
)
37、查询课程不及格的学生
-- 方法1
select s.* , c_name, s_score
from student s natural join score
natural join course
where s_score < 60
-- 方法2
select s_name, c_name, s_score
from score sc
join student s on s.s_id = sc.s_id
join course c on c.c_id = sc.c_id
where 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
-- 方法1
select s_id, s_name, s_score
from student s natural join score sc
natural join course c
where c_id = 1
and s_score >= 80
-- 方法2
select s.s_id, s_name
from student s
where s_id IN (
select s_id
from score
where c_id = 1
and s_score >= 80
)39、求每门课程的学生人数
-- 方法1
select c.c_id, c_name, count(*) cnt_student
from course c natural join score
group by c.c_id, c_name
-- 方法2
select c.c_id, c_name, cnt_student
from course c join (
select c_id, count(*) cnt_student
from score
group by c_id
) temp
on temp.c_id = c.c_id40、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
-- 方法1 张三老师所授课程中学生的最高分数， 再求张三老师所授课程中学生的成绩等于最高分数的学生
select s.*, s_score
from student s natural join score sc
natural join course c
natural join teacher t
where t_name = '张三'
and s_score = (
select max(s_score)
from student s natural join score sc
natural join course c
natural join teacher t
where t_name = '张三'
)
-- 方法2
select s.*, s_score
from student s natural join score sc
natural join course c
natural join teacher t
where t_name = '张三'
order by s_score desc limit 1
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
-- 按成绩分组，选择count(*)>1 ，再根据s_score 选择score表中的元组
select * from score
where s_score IN (
select s_score from score
group by s_score
having count(*) > 1
)42、查询每门课程成绩最好的前三名
select c.c_id, c_name, s.s_id, s_name, s_score
from (
select s_id, c_id, s_score
from score s1
where (
select count(*)
from score s2
where s2.s_score > s1.s_score
and s2.c_id = s1.c_id
) < 3
) temp
join student s on s.s_id = temp.s_id
join course c on c.c_id = temp.c_id
order by temp.c_id, temp.s_score desc43、统计每门课程的学生选修人数（超过5人的课程才统计）
select c_id, c_name , count(*) cnt_student
from score natural join course
group by c_id, c_name
having count(*)>=544、检索至少选修两门课程的学生学号
select * from student
where s_id IN (
select s_id from score
group by s_id
having count(*) >=2
)
45、查询选修了全部课程的学生信息
-- 方法1
select * from student s
where s_id IN (
select s_id from score
group by s_id
having count(*) = (
select count(*) from course
)
)
-- 方法2
select * from student
where NOT EXISTS (
select * from course
where NOT EXISTS (
select * from score
where score.c_id = course.c_id
and student.s_id = score.s_id
)
)46、查询各学生的年龄(周岁)
select *,
(
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
# 1、选取原生日里的月日成分（按照字符串进行处理）
# >>> MID(sage,6,5)
# 2、将选取的月日成分与今年的年份相接
# >>> CONCAT(YEAR(CURDATE()),"-" ,MID(sage,6,5))
# 3、使拼接后的日期的周数和 now/curdate 的周数相等即可确定本周过生日
select * from student
where week(concat(year(curdate()),"-", mid(s_birth,6,5))) = week(curdate())48、查询下周过生日的学生
select * from student
where week(concat(year(curdate()),"-", mid(s_birth,6,5))) = week(curdate())+1
49、查询本月过生日的学生
select * from student where month(s_birth) = month(now())
select * from student
where month(concat(year(curdate()), "-", mid(s_birth,6,5))) = month(curdate())
50、查询12月份过生日的学生
select * from student where month(s_birth)=12
select * from student
where month(concat(year(curdate()), "-", mid(s_birth,6,5))) = 12最后，博主把整合版本的sql练习放在下面，需要的宝子们自取。文章来源：https://www.toymoban.com/news/detail-447010.html
student-test.sql资源-CSDN文库文章来源地址https://www.toymoban.com/news/detail-447010.html
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