See Also:"Analytic Functions" for information on syntax, semantics, and restrictions
RANK calculates the rank of a value in a group of values. The return type is
See Also:Table 2-10, "Implicit Type Conversion Matrix" for more information on implicit conversion and "Numeric Precedence" for information on numeric precedence
Rows with equal values for the ranking criteria receive the same rank. Oracle Database then adds the number of tied rows to the tied rank to calculate the next rank. Therefore, the ranks may not be consecutive numbers. This function is useful for top-N and bottom-N reporting.
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
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
The following example calculates the rank of a hypothetical employee in the sample table
hr.employees with a salary of $15,500 and a commission of 5%:
SELECT RANK(15500, .05) WITHIN GROUP (ORDER BY salary, commission_pct) "Rank" FROM employees; Rank ---------- 105
Similarly, the following query returns the rank for a $15,500 salary among the employee salaries:
SELECT RANK(15500) WITHIN GROUP (ORDER BY salary DESC) "Rank of 15500" FROM employees; Rank of 15500 -------------- 4
The following statement ranks the employees in the sample
hr schema in department 60 based on their salaries. Identical salary values receive the same rank and cause nonconsecutive ranks. Compare this example with the analytic example for DENSE_RANK.
SELECT department_id, last_name, salary, RANK() OVER (PARTITION BY department_id ORDER BY salary) RANK FROM employees WHERE department_id = 60 ORDER BY RANK, last_name; DEPARTMENT_ID LAST_NAME SALARY RANK ------------- ------------------------- ---------- ---------- 60 Lorentz 4200 1 60 Austin 4800 2 60 Pataballa 4800 2 60 Ernst 6000 4 60 Hunold 9000 5