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题解 | #删除emp_no重复的记录，只保留最小的id对应的记录。#

delete from titles_test where id not in ( select * from ( select min(id) from titles_test group by emp_no ) as T1 ...

2021-08-23

0 286

题解 | #创建一个actor_name表#

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

2021-08-23

0 334

题解 | #创建一个actor_name表#

create table if not exists actor_name ( first_name varchar(45) not null, last_name varchar(45) not null ) select first_name,last_name from actor

2021-08-23

0 253

题解 | #查找描述信息中包含robot的电影对应的分类名称以及电影数目，而且还需要该分类对应电影数量>=5部#

select c.name,count(f.film_id) as count from film f left join film_category fc ON f.film_id=fc.film_id left join category c ON fc.category_id=c.cate...

2021-08-22

0 239

题解 | #汇总各个部门当前员工的title类型的分配数目#

select dp.dept_no,dp.dept_name,t.title,count(*) as count from titles t left join dept_emp de on t.emp_no=de.emp_no left join departments dp on de.dept...

2021-08-22

0 275

题解 | #获取员工其当前的薪水比其manager当前薪水还高的相关信息#

select s.emp_no,a.emp_no as manager_no,s.salary as emp_salary,a.salary as manager_salary FROM salaries sleft join dept_emp de ON s.emp_no=de.emp_nolef...

2021-08-22

0 255

题解 | #对所有员工的薪水按照salary降序进行1-N的排名#

select de.dept_no,s.emp_no,s.salary from salaries s inner join dept_emp de ON s.emp_no=de.emp_no left join dept_manager dm ON s.emp_no=dm.emp_no where...

2021-08-22

0 241

题解 | #统计各个部门的工资记录数#

select dp.dept_no,dp.dept_name,count(s.emp_no) as sum from salaries sleft join dept_emp de ON s.emp_no=de.emp_noleft join departments dp ON de.dept_no...

2021-08-22

0 282

题解 | #查找在职员工自入职以来的薪水涨幅情况#

select a.emp_no,a.salary-b.salary as growthfrom (select emp_no,salary from salaries where to_date='9999-01-01' ) ainner join (select s.emp_no ,s.salar...

2021-08-22

0 216

题解 | #查找所有员工的last_name和first_name以及对应的dept_name#

select e.last_name,e.first_name,de.dept_name from employees eleft join dept_emp d on e.emp_no=d.emp_noleft join departments de on d.dept_no=de.dept_no

2021-08-21

0 328