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将所有to_date为9999-01-01全部更新为NULL

update titles_test set to_date =NULL , from_date ="2001-01-01" where to_date = "9999-01-01"

2024-11-11

0 59

给出employees表中排名为奇数行的first_name

select first_name first from( select first_name ,row_number()over(order by first_name )rn from employees order by emp_no )t where rn % 2 =1

2024-11-11

0 49

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 id from ( select * ,row_number()over(partition by emp_no order by id asc) rn from titles_test )t w...

2024-11-11

0 53

题解 | #创建一个actor_name表#

create table actor_name as select first_name, last_name from actor

2024-11-11

0 51

题解 | 使用子查询的方式找出Action类别的电影对应信息

select title,description from film where film_id in ( select fc.film_id from film_category fc join category cy using(category_id) whe...

2024-11-09

0 47

题解 | #汇总各个部门当前员工的title类型的分配数目#

select de.dept_no,ds.dept_name,ts.title,count(1) as count from dept_emp de join departments ds using(dept_no) join titles ts using(emp_no) group...

2024-11-09

0 38

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 t1.emp_no ,t2.emp_no ,t1.salary as emp_salary ,t2.salary as manager_salary from( select emp_no ,dept_no ,salary from dept_emp join salaries u...

2024-11-09

0 41

题解 | #获取所有非manager员工当前的薪水情况#

select dp.dept_no, es.emp_no,s.salary from employees es join dept_emp dp using(emp_no) join salaries s using(emp_no) where emp_no not in (sel...

2024-11-09

0 36

题解对所有员工的薪水按照salary降序进行1-N的排名

select emp_no , salary,dense_rank()over(order by salary desc) t_rank from salaries order by t_rank ,emp_no

2024-11-09

0 36

题解 | #统计各个部门的工资记录数#

select dt.dept_no,dept_name,count(*) as 'sum' from departments dt join dept_emp dp using(dept_no) join salaries s using(emp_no) group by dt.dept_n...

2024-11-09

0 35