Oracle 进行一对一连接

Question

我想从多对多进行一对多连接。两个表格示例：

生产：

ORDER	PRODUCED_AMOUNT	PRODUCT_ID
Order1	1000	ProductID1
Order1	1000	ProductID1
Order1	5000	ProductID2

订单：

ORDER	AMOUNT_TO_PRODUCE	PRODUCT	PRODUCT_ID
Order1	600	Product1 - note1	ProductID1
Order1	600	Product1 - note2	ProductID1
Order1	600	Product1 - note3	ProductID1
Order1	5000	Product2	ProductID2

如果我编写一个示例查询，例如

SELECT ORDERS.ORDER, ORDER.AMOUNT_TO_PRODUCE, SUM(PRODUCTION.PRODUCED_AMOUNT), ORDERS.PRODUCT
FROM ORDERS JOIN PRODUCTION ON ORDERS.PRODUCT_ID = PRODUCTION.PRODUCT_ID
GROUP BY ORDERS.ORDER, ORDER.AMOUNT_TO_PRODUCE, ORDERS.PRODUCT

我明白了

ORDER	AMOUNT_TO_PRODUCE	PRODUCED_AMOUNT	PRODUCT
Order1	600	2000	Product1 - note1
Order1	600	2000	Product1 - note2
Order1	600	2000	Product1 - note3
Order1	5000	5000	Product2

但我想要

ORDER	AMOUNT_TO_PRODUCE	PRODUCED_AMOUNT	PRODUCT
Order1	600	2000	Product1 - note1
Order1	600	0	Product1 - note2
Order1	600	0	Product1 - note3
Order1	5000	5000	Product2

Answer 1

您似乎想要查找生产的总数量并将其连接到订单表中每个订单和产品 ID 的第一行。如果是这种情况，您应该：

1. 在加入生产行之前对它们求和
2. 根据订单和产品 ID 确定订单表中的第一行，您可以使用 row_number() 分析函数来完成此操作
3. 将两者连接在一起，这将涉及到订单表的总和生产行的外部连接

你可以这样做：

WITH production AS (SELECT 'Order1' order_num, 1000 produced_amount, 'ProductID1' product_id FROM dual UNION ALL
                    SELECT 'Order1' order_num, 1000 produced_amount, 'ProductID1' product_id FROM dual UNION ALL
                    SELECT 'Order1' order_num, 5000 produced_amount, 'ProductID2' product_id FROM dual),
         orders AS (SELECT 'Order1' order_num, 600 amount_to_produce, 'Product1 - note1' product, 'ProductID1' product_id FROM dual UNION ALL
                    SELECT 'Order1' order_num, 600 amount_to_produce, 'Product1 - note2' product, 'ProductID1' product_id FROM dual UNION ALL
                    SELECT 'Order1' order_num, 600 amount_to_produce, 'Product1 - note3' product, 'ProductID1' product_id FROM dual UNION ALL
                    SELECT 'Order1' order_num, 5000 amount_to_produce, 'Product2' product, 'ProductID2' product_id FROM dual),
    -- end of mimicking your tables; main query below
    prdctn_amts AS (SELECT order_num,
                           SUM(produced_amount) total_produced_amount,
                           product_id
                    FROM   production
                    GROUP BY order_num,
                             product_id),
     order_dets AS (SELECT order_num,
                           amount_to_produce,
                           product,
                           product_id,
                           row_number () OVER (PARTITION BY order_num, product_id ORDER BY product) rn
                    FROM   orders)
SELECT o.order_num,
       o.amount_to_produce,
       NVL(p.total_produced_amount, 0)
       total_produced_amount,
       o.product
FROM   order_dets o
       LEFT OUTER JOIN prdctn_amts p ON o.order_num = p.order_num
                                        AND o.product_id = p.product_id
                                        AND o.rn = 1
ORDER BY o.order_num,
         o.product_id,
         o.rn;

产生以下输出：

ORDER_NUM AMOUNT_TO_PRODUCE TOTAL_PRODUCED_AMOUNT PRODUCT
--------- ----------------- --------------------- ----------------
Order1                  600                  2000 Product1 - note1
Order1                  600                     0 Product1 - note2
Order1                  600                     0 Product1 - note3
Order1                 5000                  5000 Product2

注意您不需要 with 子句中的前两个子查询；我已经添加了它们，所以我可以模仿你的两个表，而你只需要直接引用这两个表。另外，由于ORDER 是Oracle 中的保留字，我避免了每次需要通过将列名更改为ORDER_NUM 来引用它时都必须使用" 来封装列名。