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Purpose
Calculates the rank of a value in a group of values. Rows with equal values for the ranking criteria receive the same rank.
Oracle then adds the number of tied rows to the tied rank to calculate the next rank. Therefore, the ranks may not be consecutive numbers.
As an aggregate function, RANK calculates the rank of a hypothetical row identified by the arguments of the function with respect to a given sort specification.
The arguments of the function must all evaluate to constant expressions within each aggregate group, because they identify a single row within each group.
The constant argument expressions and the expressions in the ORDER BY clause of the aggregate match by position. Therefore, the number of arguments must be the same and their types must be compatible.
As an analytic function, RANK computes the rank of each row returned from a query with respect to the other rows returned by the query, based on the values of the value_exprs in the order_by_clause.
Aggregation
Single Column RANK as Aggregation Function
RANK(<rank expression>) WITHIN GROUP
(ORDER BY <expression> <ASC|DESC> NULLS <FIRST|LAST>);
conn hr/hr@pdbdev
-- the following query returns the rank for a $15,500 salary
SELECT RANK(15500) WITHIN GROUP
(ORDER BY salary DESC NULLS LAST) SAL_RANK
FROM employees;
Multiple Column RANK as Aggregation Function
RANK(<rank expression>) WITHIN GROUP
(ORDER BY <expression> <ASC|DESC> NULLS <FIRST|LAST>);
-- the following query returns the rank of a hypothetical employee with a salary of $15,500 and commission of 3.6%
conn hr/hr@pdbdev
SELECT RANK(.36, 15500) WITHIN GROUP
(ORDER BY commission_pct, salary) RANK
FROM employees;
Multiple Column RANK as Aggregation Function
RANK(<rank expression>) WITHIN GROUP
(ORDER BY <expression> <ASC|DESC> NULLS <FIRST|LAST>);
-- the following query returns the rank of a hypothetical employee with a salary of $15,500 and commission of 3.6%
conn hr/hr@pdbdev
SELECT RANK(.36, 15500) WITHIN GROUP
(ORDER BY commission_pct NULLS FIRST, salary) RANK
FROM employees;
Analytic
With Order By Clause
RANK() OVER (<order by clause>);
-- find the employee with the 2nd highest salary
conn hr/hr@pdbdev
SELECT *
FROM (
SELECT employee_id, last_name, salary,
RANK() OVER (ORDER BY salary DESC) EMPRANK
FROM employees)
WHERE emprank = 2;
-- verify result
SELECT employee_id, last_name, salary,
RANK() OVER (ORDER BY salary DESC) EMPRANK
FROM employees;
-- find the 20 largest tables in a tablespace
conn / as sysdba
col segment_name format a30
SELECT * FROM (
SELECT segment_name, blocks, RANK() OVER(ORDER BY blocks DESC) SZ
FROM dba_segments
WHERE segment_type = 'TABLE'
AND TABLESPACE_NAME = 'UWDATA')
WHERE SZ < 21;
With Partition By Clause
RANK() OVER (<query partition clause> <order by clause>);
conn hr/hr@pdbdev
SELECT department_id,last_name,salary,commission_pct,
RANK() OVER (PARTITION BY department_id ORDER BY salary DESC, commission_pct) RANK
FROM employees;
SELECT department_id,last_name,salary,commission_pct,
RANK() OVER (PARTITION BY department_id ORDER BY salary DESC, commission_pct) RANK
FROM employees
WHERE department_id = 80;
