php - 比较同一个表中的两个值

Question

我有一个表，我想获取其所有名称工资低于下一个人的人同一张 table ，我已经尝试过这个但是它不起作用有什么建议吗？

select t1.Name, t2.Name as Name2 
from employees t1 
inner join employees t2 on 
t1.ID = t2.ID
where t1.Salary < t2.Salary;

我正在尝试打印出每个工资低于下一个的人的姓名

Joe "has less then" Bob
Joe "has less then" foo
Joe "has less then" Bar
Joe "has less then" Pete

Answer 1

比较当前和下一个 ID 列

select t1.Name, t2.Name as Name2 ,t1.salary,t2.salary
from employees t1 
inner join employees t2 on 
t1.ID+1 = t2.Id
where t1.Salary < t2.Salary;