RANK
The RANK() function assigns a unique rank to each value within an ordered group of values.
The rank value starts at 1 and continues up sequentially. If two values are the same, they have the same rank.
Analyze Syntax
func.rank().over(partition_by=[<columns>], order_by=[<columns>])
Analyze Examples
table.employee_id, table.first_name, table.last_name, table.department, table.salary, func.rank().over(order_by=table.salary).alias('rank')
| employee_id | first_name | last_name | department | salary | rank |
|-------------|------------|-----------|------------|--------|------|
| 1 | John | Doe | IT | 90000 | 1 |
| 2 | Jane | Smith | HR | 85000 | 2 |
| 3 | Mike | Johnson | IT | 82000 | 3 |
| 4 | Sara | Williams | Sales | 77000 | 4 |
| 5 | Tom | Brown | HR | 75000 | 5 |
SQL Syntax
RANK() OVER (
[ PARTITION BY <expr1> ]
ORDER BY <expr2> [ { ASC | DESC } ]
[ <window_frame> ]
)
SQL Examples
Create the table
CREATE TABLE employees (
employee_id INT,
first_name VARCHAR,
last_name VARCHAR,
department VARCHAR,
salary INT
);
Insert data
INSERT INTO employees (employee_id, first_name, last_name, department, salary) VALUES
(1, 'John', 'Doe', 'IT', 90000),
(2, 'Jane', 'Smith', 'HR', 85000),
(3, 'Mike', 'Johnson', 'IT', 82000),
(4, 'Sara', 'Williams', 'Sales', 77000),
(5, 'Tom', 'Brown', 'HR', 75000);
Ranking employees by salary
SELECT
employee_id,
first_name,
last_name,
department,
salary,
RANK() OVER (ORDER BY salary DESC) AS rank
FROM
employees;
Result:
employee_id | first_name | last_name | department | salary | rank |
---|---|---|---|---|---|
1 | John | Doe | IT | 90000 | 1 |
2 | Jane | Smith | HR | 85000 | 2 |
3 | Mike | Johnson | IT | 82000 | 3 |
4 | Sara | Williams | Sales | 77000 | 4 |
5 | Tom | Brown | HR | 75000 | 5 |
Last modified June 11, 2024 at 7:47 PM EST: adding window functions (6bcb2f2)