KEEP
函数用于对一组行中的值进行操作,将这组行按照给定的排序规则排序后返回排在第一或最后的值。作为聚合函数,KEEP
对所有行进行操作并返回单个输出行。作为分析函数,KEEP
基于 query_partition_clause
中的一个或多个表达式将查询结果集分为几组。
KEEP
函数必须与 MIN
、MAX
、SUM
、AVG
、COUNT
、VARIANCE
或 STDDEV
函数一起使用。
语法
KEEP (DENSE_RANK {FIRST | LAST} ORDER BY expr [ DESC | ASC ] [ NULLS { FIRST | LAST } ] [, expr [ DESC | ASC ] [ NULLS { FIRST | LAST } ]]... )
[ OVER ( [query_partition_clause] ) ]
作为分析函数使用时,您需要使用窗口函数的完整语法,它对一组行的集合进行计算并返回多个值。作为聚合函数使用时,该函数对一组行的集合进行聚合计算,结果只能返回一个值,此时不需要加 OVER
关键字。
参数
参数 | 说明 |
---|---|
OVER | 使用 |
expr | 可以是任何表达式。度量列中的空值将被忽略。 |
返回类型
返回与度量列相同的数据类型。
示例
分析函数示例
建表 employees,并向里面插入数据,执行以下语句:
CREATE TABLE employees (department_id INT,manager_id INT,last_name varchar(50),hiredate varchar(50),SALARY INT);
INSERT INTO employees VALUES(30, 100, 'Raphaely', '2017-07-01', 1700);
INSERT INTO employees VALUES(30, 100, 'De Haan', '2018-05-01',11000);
INSERT INTO employees VALUES(40, 100, 'Errazuriz', '2017-07-21', 1400);
INSERT INTO employees VALUES(50, 100, 'Hartstein', '2019-10-05',14000);
INSERT INTO employees VALUES(50, 100, 'Raphaely', '2017-07-22', 1700);
INSERT INTO employees VALUES(50, 100, 'Weiss', '2019-10-05',13500);
INSERT INTO employees VALUES(90, 100, 'Russell', '2019-07-11', 13000);
INSERT INTO employees VALUES(90,100, 'Partners', '2018-12-01',14000);
对于每个部门:按照薪水排序,对排在第一的员工的薪水求和;按照雇用日期排序,对排在最后的员工的薪水求和。执行以下语句:
SELECT last_name, department_id, salary,
SUM(salary) KEEP (DENSE_RANK FIRST ORDER BY salary)
OVER (PARTITION BY department_id) "Worst",
SUM(salary) KEEP (DENSE_RANK LAST ORDER BY HIREDATE)
OVER (PARTITION BY department_id) "Best"
FROM employees
ORDER BY department_id, salary, last_name;
查询结果如下:
+-----------+---------------+--------+-------+-------+
| LAST_NAME | DEPARTMENT_ID | SALARY | Worst | Best |
+-----------+---------------+--------+-------+-------+
| Raphaely | 30 | 1700 | 1700 | 11000 |
| De Haan | 30 | 11000 | 1700 | 11000 |
| Errazuriz | 40 | 1400 | 1400 | 1400 |
| Raphaely | 50 | 1700 | 1700 | 27500 |
| Weiss | 50 | 13500 | 1700 | 27500 |
| Hartstein | 50 | 14000 | 1700 | 27500 |
| Russell | 90 | 13000 | 13000 | 13000 |
| Partners | 90 | 14000 | 13000 | 13000 |
+-----------+---------------+--------+-------+-------+
8 rows in set (0.01 sec)
聚合函数示例
建表 employees,并向里面插入数据,执行以下语句:
CREATE TABLE employees (department_id INT,manager_id INT,last_name varchar(50),hiredate varchar(50),SALARY INT);
INSERT INTO employees VALUES(30, 100, 'Raphaely', '2017-07-01', 1700);
INSERT INTO employees VALUES(30, 100, 'De Haan', '2018-05-01',11000);
INSERT INTO employees VALUES(40, 100, 'Errazuriz', '2017-07-21', 1400);
INSERT INTO employees VALUES(50, 100, 'Hartstein', '2019-10-05',14000);
INSERT INTO employees VALUES(50, 100, 'Raphaely', '2017-07-22', 1700);
INSERT INTO employees VALUES(50, 100, 'Weiss', '2019-10-05',13500);
INSERT INTO employees VALUES(90, 100, 'Russell', '2019-07-11', 13000);
INSERT INTO employees VALUES(90,100, 'Partners', '2018-12-01',14000);
按照薪水排序,对排在第一的员工的薪水求和;按照雇用日期排序,对排在最后的员工的薪水求和。执行以下语句:
SELECT
SUM(salary) KEEP (DENSE_RANK FIRST ORDER BY salary) "Worst",
SUM(salary) KEEP (DENSE_RANK LAST ORDER BY HIREDATE) "Best"
FROM employees;
查询结果如下:
+-------+-------+
| Worst | Best |
+-------+-------+
| 1400 | 27500 |
+-------+-------+
1 row in set (0.00 sec)
作者:唐伯虎点蚊香,如若转载,请注明出处:https://www.web176.com/oceanbase/26281.html