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题解 | #牛客每个人最近的登录日期(五)#

写的无敌麻烦写完借鉴下评论区各位大神的还是把自己也上传了吧#sql第70 selectc.date,round(ifnull(count(d.user_id),0),3)from( select distinct date from login) cleft join ( ...

2021-07-31

0 436

题解 | #牛客每个人最近的登录日期(四)#

黔驴技穷了，主要对空值函数不熟悉selectm,count(user_id)from ( select user_id ,min(date) m from login group by user_id) agroup by a.m#结果正确但不带没有日期的值 sele...

2021-07-31

0 399

题解 | #牛客每个人最近的登录日期(二)#

selecta.name1,a.name2,b.mfrom ( select l.* ,u.name name1 ,c.name name2 from login l join user u on l.user_id =u.id join client...

2021-07-30

0 394

题解 | #异常的邮件概率#

selecta.date,round(count(type)/a.r1,3)from ( select e.* ,count(type) over(partition by date) r1 from email e join user u1 on e.send_...

2021-07-30

1 465

题解 | #对于employees表中，给出奇数行的first_name#

selectfirst_namefrom ( select first_name ,row_number() over(order by first_name) r1 from employees) awhere mod(r1,2) = 1吐了题目说不用排序输出后有不...

2021-07-28

0 578

题解 | #获取有奖金的员工相关信息。#

selecte1.emp_no,e1.first_name,e1.last_name,eb1.btype,s1.salary,s1.salary0.1 bonusfrom employees e1join emp_bonus eb1on e1.emp_no = eb1.emp_nojoin sala...

2021-07-28

0 643

题解 | #查找排除当前最大、最小salary之后的员工的平均工资avg_salary#

SQL 第54题解法一selectavg(salary)from salarieswhere salary not in ( select min(salary) from salaries where to_date = '9999-01-01')and salary n...

54 翻车了很多遍

2021-07-27

35 1304

题解 | #将所有获取奖金的员工当前的薪水增加10%#

update salaries set salary = salary*1.1 where emp_no in ( select emp_no from emp_bonus ) and to_date = "999...

#sql

2021-07-26

0 501

题解 | #创建一个actor_name表#

create table actor_name( first_name varchar(45) NOT NULL, last_name varchar(45) NOT NULL );insert into actor_name(first_name,last_name ) s...

2021-07-24

0 324

题解 | #创建一个actor表，包含如下列信息#

create table actor ( actor_id smallint(5) not null, first_name varchar(45) not null, last_name varchar(45) not null, la...

2021-07-24

0 431