在'PARTITION BY'上使用过滤条件

Question

我有两个表。一个是Reference表，用于排序优先级，一个是Customer表。 Reference表用于优先考虑表Customer中的每个列，以便为单个客户的单个列给出不同的顺序。

参考表：

--------------------------------------- 
| Priority | Attribute | sourceID | 
--------------------------------------- 
| 1  |  EMAIL |  1  | 
| 2  |  EMAIL |  2  | 
| 3  |  EMAIL |  3  | 
| 2  |  NAME |  1  | 
| 1  |  NAME |  2  | 
| 3  |  NAME |  3  | 
--------------------------------------- 

客户表：

----------------------------------------------------------------------- 
| CustomerID | Name |  Email  | SourceID |  Date | 
----------------------------------------------------------------------- 
| 1  | John |  NULL   |  1  | 03/01/2017 | 
| 1  | NULL | John@email.com |  3  | 01/01/2017 | 
| 1  | J  | J.Smith@email.com |  2  | 02/01/2017 | 
----------------------------------------------------------------------- 

结果：

--------------------------------------------- 
| CustomerID | Name |  Email  | 
--------------------------------------------- 
|  1  | John | J.Smith@email.com | 
--------------------------------------------- 

在t他目前我使用下面的查询做到这一点：

SELECT DISTINCT 
     FIRST_VALUE(c.Name IGNORE NULLS) 
      OVER (PARTITION BY p.customerID 
       ORDER BY r.PRIORITY, c.DATE 
       ROWS BETWEEN UNBOUNDED PRECEDING AND UNBOUNDED FOLLOWING) AS EMAIL, 
     FIRST_VALUE(c.Email IGNORE NULLS) 
      OVER (PARTITION BY c.customerID 
       ORDER BY r.PRIORITY, c.DATE 
       ROWS BETWEEN UNBOUNDED PRECEDING AND UNBOUNDED FOLLOWING) AS EMAIL 
FROM Customer c 
    JOIN reference r ON c.sourceID = r.sourceID; 

然而，这确实需要为每列不同的属性纳入考虑。我需要按部分为每个分区添加某种过滤器。

任何人都可以协助我如何去做到这一点？

Answer 1

一种方法是把属性为客户在一列，然后重新组合它们：

SELECT DISTINCT customerId 
     first_value(CASE WHEN ca.attribute = 'NAME' THEN ca.val end) OVER 
      (PARTITION BY ca.customerId, attribute ORDER BY r.priority, ca.date) AS name, 
     first_value(CASE WHEN ca.attribute = 'EMAIL' THEN ca.val END) OVER 
      (PARTITION BY ca.customerId, attribute ORDER BY r.priority, ca.date) AS email 
FROM ((SELECT customerId, 'NAME' AS attribute, name AS val, sourceId, date 
     FROM customer c 
    ) UNION ALL 
     (SELECT customerId, 'EMAIL' AS attribute, email AS val, sourceId, date 
     FROM customer c 
    ) 
    ) ca JOIN 
    reference r 
    ON r.sourceId = ca.sourceId AND r.attribute = ca.attribute; 

注意，这里使用SELECT DISTINCT而不是GROUP BY。我不认为Netezza具有聚合函数first_value()，所以这个构造解决了这个问题。