SQL 分析函数：rank() over partition by not working property

Posted 2023-03-29

【中文标题】SQL 分析函数：rank() over partition by not working property

【英文标题】：SQL analytical function: rank() over partition by not working property

【发布时间】：2015-03-24 20:38:32

【问题描述】：

CREATE TABLE customers
( customer_id number(10) not null,
  customer_name varchar2(50) not null
);

INSERT INTO customers VALUES(22,'W');
INSERT INTO customers VALUES(22,'W');
INSERT INTO customers VALUES(20,'Q');
INSERT INTO customers VALUES(20,'Q');
COMMIT;

现在我正在尝试按客户名称获取与我的分区相对应的不同等级

SELECT DENSE_RANK() OVER(PARTITION BY customer_name ORDER BY CUSTOMER_ID) , CUSTOMER_ID FROM CUSTOMERS;

输出：

1   20
1   20
1   22
1   22

预期输出：

1   20
1   20
2   22
2   22

【参考方案1】：

使用下面的查询

SELECT 
DENSE_RANK() OVER(ORDER BY CUSTOMER_ID) , 
CUSTOMER_ID 
FROM CUSTOMERS;

删除 partition BY by 子句，因为如果您使用 partition by 子句，它将根据名称划分结果。