推荐一款inscode内的模板SQL练习生，此文附带目前所有题的答案
如有错误欢迎斧正~

https://inscode.csdn.net/@TPEngineer/SQLBoy

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个人总结的答案（已涵盖目前的所有题，题库更新踢我）

练习

1.基础语法 - 查询 - 全表查询

-- 请在此处输入 SQL
select * from student;

2.基础语法 - 查询 - 选择查询

-- 请在此处输入 SQL
select name,age from student;

主线关卡

1.基础语法 - 查询 - 全表查询

-- 请在此处输入 SQL
select * from student;

2.基础语法 - 查询 - 选择查询

-- 请在此处输入 SQL
select name,age from student;

3.基础语法 - 查询 - 别名

-- 请在此处输入 SQL
select name as 学生姓名,
age as 学生年龄
from student;

4.基础语法 - 查询 - 常量和运算

-- 请在此处输入 SQL
select name,score,2*score as double_score from student;

5.基础语法 - 条件查询 - where

-- 请在此处输入 SQL
select name,score 
from student
where name='鱼皮';

6.基础语法 - 条件查询 - 运算符

-- 请在此处输入 SQL
select name,age 
from student
where name!='热dog'

7.基础语法 - 条件查询 - 空值

-- 请在此处输入 SQL
select name,age,score 
from student
where age is not null;

8.基础语法 - 条件查询 - 模糊查询

-- 请在此处输入 SQL
select name,score 
from student
where name not like '%李%';

9.基础语法 - 条件查询 - 逻辑运算

-- 请在此处输入 SQL
select name,score 
from student
where name like '%李%'  or score>500;

10.基础语法 - 去重

-- 请在此处输入 SQL
select distinct class_id,exam_num 
from student;

11.基础语法 - 排序

-- 请在此处输入 SQL
select name,age,score 
from student
order by score desc,
age asc;

12.基础语法 - 截断和偏移

-- 请在此处输入 SQL
select name,age from student
order by age asc
limit 1,3;

13.基础语法 - 条件分支

-- 请在此处输入 SQL
select name,
    case when (age>60) then '老同学'
         when (age>20) then '年轻'
         else '小同学' end as age_level
from student
order by name asc;

14.函数 - 时间函数

-- 请在此处输入 SQL
select name,
date() as '当前日期' 
from student;

15.函数 - 字符串处理

-- 请在此处输入 SQL
select id ,
name ,
upper(name) as upper_name 
from student where name=='热dog';

16.函数 - 聚合函数

-- 请在此处输入 SQL
select sum(score) as total_score,
AVG(score) as avg_score,
MAX(score) as max_score,
MIN(score) as min_score from student;

17.分组聚合 - 单字段分组

-- 请在此处输入 SQL
select class_id,
AVG(score) as avg_score 
from student 
group by class_id;

18.分组聚合 - 多字段分组

-- 请在此处输入 SQL
select class_id,
exam_num,
Count(id) as total_num 
from student 
group by class_id,exam_num;

19.分组聚合 - having 子句

-- 请在此处输入 SQL
select class_id,
SUM(score) as total_score 
from student
group by class_id
having SUM(score)>150;

20.查询进阶 - 关联查询 - cross join

-- 请在此处输入 SQL
select s.name as student_name,
s.age as student_age,
s.class_id as class_id,
c.name as class_name
from student s 
cross join
class c;

21.查询进阶 - 关联查询 - inner join

-- 请在此处输入 SQL
select s.name as student_name,
s.age as student_age,
s.class_id as class_id,
c.name as class_name,
c.level as class_level
from student s
join class c on s.class_id=c.id;

22.查询进阶 - 关联查询 - outer join

-- 请在此处输入 SQL
select s.name as student_name,
s.age as student_age,
s.class_id as class_id,
c.name as class_name,
c.level as class_level
from student s
left join class c
on s.class_id=c.id;

23.查询进阶 - 子查询

-- 请在此处输入 SQL
select 
    s.name,
    s.score,
    s.class_id
from 
    student s
where 
    s.class_id IN (SELECT  c.id FROM class c);

24.查询进阶 - 子查询 - exists

-- 请在此处输入 SQL
select name,age,class_id 
from student
where not exists(
    select 1
    from class
    where class.id==student.class_id
);

25.查询进阶 - 组合查询

-- 请在此处输入 SQL
select name,age,score,class_id 
from student
union all
select name,age,score,class_id
from student_new;

26.查询进阶 - 开窗函数 - sum over

-- 请在此处输入 SQL
select id,name,age,score,class_id,
avg(score) over (partition by class_id)
as class_avg_score 
from student;

27.查询进阶 - 开窗函数 - sum over order by

-- 请在此处输入 SQL
select id,name,age,score,class_id,
sum(score) over (partition by class_id 
order by score asc)
as class_sum_score
from student;

28.查询进阶 - 开窗函数 - rank

-- 请在此处输入 SQL
select id,name,age,score,class_id,
rank() over 
(partition by class_id order by score desc)
as ranking
from student;

29.查询进阶 - 开窗函数 - row_number

-- 请在此处输入 SQL
select id,name,age,score,class_id,
row_number() over
(partition by class_id order by score desc)
as row_number
from student;

30.查询进阶 - 开窗函数 - lag / lead

-- 请在此处输入 SQL
select id,name,age,score,class_id,
lag(name,1,null) over 
(partition by class_id order by score desc)
as prev_name,
lead(name,1,null) over 
(partition by class_id order by score desc)
as next_name
from student;

自定义关卡

冒险者和金币

-- 请在此处输入 SQL
select adventurer_id,adventurer_name,
sum(reward_coins) as total_reward_coins
from rewards
group by adventurer_id,adventurer_name
order by total_reward_coins desc
limit 3;

魔法学院

-- 请在此处输入 SQL
select student_id,student_name,subject_id,subject_name,score,
rank() over (partition by subject_id order by score desc)
as score_rank
from magic_scores;

大浪淘鸡

-- 请在此处输入 SQL
select observer_name,observation_date,wave_intensity from chicken_observation
where observation_location like '%大浪淘鸡%'
and wave_intensity >5;