The challenge is to find the employees with the second highest salary in each department. However, it is a little more complicated because if two employees have the same salary, you need to list both of them.
Solution to TSQL Beginners Challenge 1
;with cte as (
select EmployeeID, EmployeeName, Department, dense_rank() over (partition by department order by salary desc ) as 'rank', salary
select EmployeeID, EmployeeName, Department, Salary from cte
where [RANK] = 2;