SELECT MIN(EmployeeSalary)
FROM (SELECT DISTINCT TOP (2) EmployeeSalary
FROM Employee
ORDER BY EmployeeSalary DESC)T
HAVING MIN(EmployeeSalary) <> MAX(EmployeeSalary);

This soln is only for 2nd highest.

select max(salary) from employee where salary <(select max(salary) from employee)

for nth highest replace n with any value:

select top 1 salary from (
select top N salary from employee order by salary desc) result
order by salary

another one:

with cte as(
select salary,dense_rank() over (order by salary desc) rank from employee )
select salary from cte where rank = n;

Follow for more solutions.

SELECT TOP 1 EmployeeSalary
FROM (
    SELECT DISTINCT TOP 2 EmployeeSalary
    FROM Employee
    ORDER BY EmployeeSalary DESC
) AS Salary
ORDER BY EmployeeSalary ASC;

1. We start by selecting the top 2 unique salary values from the Employee table in descending order, using the DISTINCT and TOP clauses, respectively.
2. We wrap this query in a subquery and alias it as Salary.
3. We select the top 1 salary value from the subquery in ascending order using the TOP and ORDER BY clauses.

To get the second highest salary from the Employee table, you can use the SELECT TOP clause along with the ORDER BY clause in SQL.