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全部文章 / 题解（共34篇）

题解 | #牛客的课程订单分析(三)#

select o.* from order_info o where user_id in (select distinct user_id from order_info where date > '2025-10-15' ...

Mysql

2021-10-05

0 320

题解 | #牛客的课程订单分析(二)#

select user_id from order_info where date > '2025-10-15' and status = 'completed' and product_name in("C++","Java","Python") group by user_id havi...

Mysql

2021-10-05

0 333

题解 | #牛客的课程订单分析(一)#

select id,user_id,product_name,status,client_id,date from order_info where status = "completed" and date > '2025-10-15' and (product_name in("C++",...

Mysql

2021-10-05

0 288

题解 | #统计salary的累计和running_total#

select s1.emp_no,s1.salary,sum(s2.salary) running_total from salaries s1 join salaries s2 on s1.emp_no >= s2.emp_no where s1.to_date = '9999-01-01...

Mysql

2021-10-05

0 275

题解 | #获取有奖金的员工相关信息。#

select eb.emp_no,e.first_name,e.last_name,eb.btype,s.salary, (case eb.btype when 1 then s.salary*0.1 when 2 then s.salary*0.2 ...

Mysql

2021-10-05

2 350

题解 | #平均工资#

select avg(salary) avg_salary from salaries where salary not in(select max(salary) from salaries where to_date = '9999-01-01') and salary not in(selec...

Mysql

2021-10-05

0 274

题解 | #在audit表上创建外键约束，其emp_no对应employees_test表的主键id#

alter table audit add CONSTRAINT hhh foreign key (emp_no) references employees_test(id);

Mysql

2021-10-05

0 300

题解 | #对first_name创建唯一索引uniq_idx_firstname#

create unique index uniq_idx_firstname on actor(first_name); create index idx_lastname on actor(last_name);

Mysql

2021-10-05

0 249

题解 | #使用子查询的方式找出属于Action分类的所有电影对应的title,description#

select title,DESCRIPTION from film where film_id in (select fc.film_id from film_category fc join category c on fc.category_id = c.categ...

Mysql

2021-10-05

0 252

题解 | #使用join查询方式找出没有分类的电影id以及名称#

select fm.film_id,fm.title from film fm left join film_category fc on fc.film_id = fm.film_id where fc.category_id is null

Mysql

2021-10-05

0 248