Purpose
MAX() returns the maximum value in the specified column. You can use this function as an aggregate or analytic function.
Note
- When you use this function as an analytic function, use the
OVERclause to define a window over the data on which the function operates. The function operates on a group of rows to return a list of values. - When you use this function as an aggregate function, the function operates on a set of rows and returns a single value. You do not need to add the
OVERclause.
Syntax
MAX ([ DISTINCT | UNIQUE | ALL ] expr) [ OVER (analytic_clause) ]
Parameters
| Parameter | Description |
|---|---|
| DISTINCT | UNIQUE | ALL | Specifies whether to remove duplicates during the execution of the query. This parameter is optional. Default value: ALL.
|
| expr | An expression or column of the numeric, character, date, or other types. |
| OVER | You can use the OVER clause to define a window over the data on which the function operates. For more information, see Analytic functions. |
Return type
The return type is the same as the data type of the expr parameter.
Examples
Assume that you have created the table employees.
obclient> SELECT * FROM employees;
+---------------+------------+-----------+------------+--------+
| DEPARTMENT_ID | MANAGER_ID | LAST_NAME | HIREDATE | SALARY |
+---------------+------------+-----------+------------+--------+
| 30 | 100 | Raphaely | 2017-07-01 | 1700 |
| 30 | 100 | De Haan | 2018-05-01 | 11000 |
| 40 | 100 | Errazuriz | 2017-07-21 | 1400 |
| 50 | 100 | Hartstein | 2019-10-05 | 14000 |
| 50 | 100 | Raphaely | 2017-07-22 | 1700 |
| 50 | 100 | Weiss | 2019-10-05 | 13500 |
| 90 | 100 | Russell | 2019-07-11 | 13000 |
| 90 | 100 | Partners | 2018-12-01 | 14000 |
+---------------+------------+-----------+------------+--------+
8 rows in set
Example of an aggregate function
Query for the maximum value in the salary column.
obclient> SELECT MAX(salary) FROM employees;
+-------------+
| MAX(SALARY) |
+-------------+
| 14000 |
+-------------+
1 row in set
Example of an analytic function
Group the data records in the table by the department_id column. Return the maximum value in the salary column.
obclient> SELECT department_id,last_name,salary,
MAX(salary) OVER (PARTITION BY department_id) AS rmax_sal
FROM employees;
+---------------+-----------+--------+----------+
| DEPARTMENT_ID | LAST_NAME | SALARY | RMAX_SAL |
+---------------+-----------+--------+----------+
| 30 | Raphaely | 1700 | 11000 |
| 30 | De Haan | 11000 | 11000 |
| 40 | Errazuriz | 1400 | 1400 |
| 50 | Hartstein | 14000 | 14000 |
| 50 | Raphaely | 1700 | 14000 |
| 50 | Weiss | 13500 | 14000 |
| 90 | Russell | 13000 | 14000 |
| 90 | Partners | 14000 | 14000 |
+---------------+-----------+--------+----------+
8 rows in set