MySQL高阶1965-丢失信息的雇员

目录

题目

准备数据

分析数据

实现

题目

编写解决方案，找到所有 丢失信息 的雇员 id。当满足下面一个条件时，就被认为是雇员的信息丢失：

• 雇员的 姓名 丢失了，或者
• 雇员的 薪水信息 丢失了

返回这些雇员的 id  employee_id ， 从小到大排序 。

准备数据

Create table If Not Exists Employees (employee_id int, name varchar(30))
Create table If Not Exists Salaries (employee_id int, salary int)
    Truncate table Employees
    insert into Employees (employee_id, name) values ('2', 'Crew')
    insert into Employees (employee_id, name) values ('4', 'Haven')
    insert into Employees (employee_id, name) values ('5', 'Kristian')
    Truncate table Salaries
    insert into Salaries (employee_id, salary) values ('5', '76071')
    insert into Salaries (employee_id, salary) values ('1', '22517')
    insert into Salaries (employee_id, salary) values ('4', '63539')

Salaries表

Employees表

 分析数据

第一步:通过union all对两个select语句进行连接,并通过左连接和右连接将不符合的数据为null

select e.Employee_id,e.name,s.salary
from Employees e left join Salaries s on e.employee_id=s.employee_id
union all
select s1.employee_id,e1.name,s1.salary
from Employees e1 right join Salaries s1 on e1.employee_id=s1.employee_id;

第二步:选出符合条件的,并且进行排序

with a as
         (select e.Employee_id,e.name,s.salary
          from Employees e left join Salaries s on e.employee_id=s.employee_id
          union all
          select s1.employee_id,e1.name,s1.salary
          from Employees e1 right join Salaries s1 on e1.employee_id=s1.employee_id)
select employee_id
from a
where name is null or salary is null
order by employee_id;

实现

with a as
         (select e.Employee_id,e.name,s.salary
          from Employees e left join Salaries s on e.employee_id=s.employee_id
          union all
          select s1.employee_id,e1.name,s1.salary
          from Employees e1 right join Salaries s1 on e1.employee_id=s1.employee_id)
select employee_id
from a
where name is null or salary is null
order by employee_id;