How can you find the second highest salary from a table?

To find the second highest salary from a table in SQL, there are multiple methods, depending on what your SQL dialect supports. Here are the most common approaches:

✅ Method 1: Using LIMIT and OFFSET (MySQL, PostgreSQL)

SELECT DISTINCT salary
FROM employees
ORDER BY salary DESC
LIMIT 1 OFFSET 1;

• LIMIT 1 OFFSET 1 skips the highest salary and returns the next one.
• DISTINCT ensures no duplicates if multiple employees have the same salary.

✅ Method 2: Using a Subquery with MAX()

SELECT MAX(salary) AS second_highest_salary
FROM employees
WHERE salary < (SELECT MAX(salary) FROM employees);

• This returns the maximum salary that is less than the highest one, i.e., the second highest.
• Works in most SQL dialects (MySQL, PostgreSQL, SQL Server, Oracle, etc.)

✅ Method 3: Using DENSE_RANK() (SQL Server, PostgreSQL, Oracle)

SELECT salary
FROM (
    SELECT salary, DENSE_RANK() OVER (ORDER BY salary DESC) AS rank
    FROM employees
) ranked
WHERE rank = 2;

• This handles ties properly — e.g., if two people have the highest salary, the next distinct one is ranked 2nd.

✅ Method 4: Using TOP with Subquery (SQL Server)

SELECT TOP 1 salary
FROM employees
WHERE salary < (
    SELECT MAX(salary) FROM employees
)
ORDER BY salary DESC;

• This is similar to Method 2 but written using SQL Server syntax.

Which One Should You Use?

Let me know if you want to get the employee(s) with the second highest salary too — I can show that as well.