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全部文章（共26篇）

题解 | 最长连续登录天数

select user_id,max(连续登陆天数) max_consec_days from (select user_id,count(差值) 连续登陆天数 from (select *,day(fdate)-row_number()over(partition by user_id order...

2025-03-09

0 57

题解 | #牛客每个人最近的登录日期(三)#

#先拿出user_id，date这两列去重 #再用窗口函数找出首次登录后下一次登录的日期，计算二者差值是否为1，若为1，就留存成功了 with t1 as ( select distinct user_id, date ...

2024-09-10

0 73

题解 | #牛客每个人最近的登录日期(二)#

select person_name, client_name, date from ( select t2.name person_name, t3.name client_name, ...

2024-09-08

0 83

题解 | #异常的邮件概率#

select distinct date, round(ration, 3) from ( select date, type, send_id, receive...

2024-09-08

0 103

题解 | #给出employees表中排名为奇数行的名#

select first_name from ( select emp_no, first_name, row_number() over ( order by ...

2024-09-08

0 92

题解 | #获取有奖金的员工相关信息。#

select t1.emp_no en, first_name, last_name, btype, salary, case when btype = 1 then salary * 0.1 when btype = ...

2024-09-08

0 127

题解 | #获取员工当前的薪水比其manager当前薪水还高

#将员工信息和工资表连接起来 # select t1.emp_no e1,t1.dept_no d1,t2.dept_no d2,t2.emp_no e2,salary s # from dept_emp t1 left join dept_manager t2 on (t1.dept_no=t2....

2024-09-03

0 108

题解 | #查找在职员工自入职以来的薪水涨幅情况#

# #最开始工资 # select employees.emp_no e1,salary s1,from_date f1,to_date t2 # from employees left join salaries on (employees.emp_no=salaries.emp_no) # wh...

2024-09-02

1 101

题解 | #获取当前薪水第二多的员工的薪水salary#

#先寻找最高工资 select max(salary) from salaries #再寻找第二高工资 # select max(salary) # from salaries # where salary != (select max(salary) # from salaries) #找到工资...

2024-09-01

0 96

题解 | #获取当前薪水第二多的员工信息#

select emp_no,salary from salaries where salary = (select salary from salaries group by salary order by salary desc limit 1 offset 1)

2024-09-01

0 92