Complete the code to assign a rank to each employee based on their salary using RANK().
SELECT employee_id, salary, RANK() OVER (ORDER BY salary [1]) AS salary_rank FROM employees;The RANK() function assigns ranks starting from the lowest salary when ordered ascending (ASC).
Complete the code to assign dense ranks to employees based on their salary using DENSE_RANK().
SELECT employee_id, salary, DENSE_RANK() OVER (ORDER BY salary [1]) AS dense_salary_rank FROM employees;DENSE_RANK() assigns ranks without gaps when ordering salaries ascending.
Fix the error in the query that tries to rank employees by salary but uses an invalid keyword.
SELECT employee_id, salary, RANK() OVER (ORDER BY salary [1]) AS rank FROM employees;The ORDER BY clause inside OVER() must use ASC or DESC, not GROUP BY or FILTER.
Fill both blanks to create a query that shows both RANK and DENSE_RANK for employees ordered by salary.
SELECT employee_id, salary, RANK() OVER (ORDER BY salary [1]) AS rank, DENSE_RANK() OVER (ORDER BY salary [2]) AS dense_rank FROM employees;
Both RANK() and DENSE_RANK() should order salaries ascending (ASC) to compare ranks properly.
Fill all three blanks to create a query that ranks employees by salary descending, shows dense ranks, and filters for ranks less than 4.
SELECT * FROM (SELECT employee_id, salary, RANK() OVER (ORDER BY salary [1]) AS rank, DENSE_RANK() OVER (ORDER BY salary [2]) AS dense_rank FROM employees) AS sub WHERE rank [3] 4;
Ordering by salary descending (DESC) ranks highest salaries first. Filtering ranks less than 4 shows top 3 ranks.