你不需要任何加入。获得所有经理的最高工资，并检查员工工资是否大于经理的最高工资

select a.Name
from Employee a 
where a.Salary > 
(SELECT MAX(salary) FROM Employee m WHERE m.id IN (SELECT managerID FROM Employee))--or however you determine managers

您可以使用EXISTS来代替：

SELECT E.[Name]
FROM dbo.Employee E
WHERE EXISTS (SELECT 1
              FROM dbo.Employee E2
                   JOIN dbo.Employee M ON E2.ManagerID = M.id
              WHERE M.Salary < E.Salary);

不过EXISTS有点乱，因为判断某人是否是经理的唯一方法是某人是否让他们担任经理。

如果要使用联接而不是子查询：

SELECT DISTINCT a.Name
FROM Employee a
CROSS JOIN Employee b 
WHERE a.Salary > b.Salary and b.Manager ID IS NULL

获得比任何经理都多的
表示您希望员工的收入高于经理的最低薪金：

select name from employees 
where salary > ( 
  select min(salary) from employees
  where id in (select distinct managerid from employees)
)

或

select name from employees 
where salary > ANY ( 
  select salary from employees
  where id in (select distinct managerid from employees)
)

如果只希望非经理的员工使用CTE：

with cte as (select distinct managerid from employees)
select name from employees 
where 
  id not in cte
  and 
  salary > (select min(salary) from employees where id in cte)

SELECT *
FROM employees e,
  (SELECT *
   FROM employees
   WHERE emp_id IN
       (SELECT manager_id
        FROM employees)) a
WHERE e.salary >a.salary
  AND e.manager_id = a.emp_id;