交叉连接剩余组合
Cross join remaining combinations
我正在尝试构建一个 table，它将结合我可以销售的所有产品，基于当前的产品。
产品状态Table
+-------------+--------------+----------------+
| customer_id | product_name | product_status |
+-------------+--------------+----------------+
| 1 | A | Active |
| 2 | B | Active |
| 2 | C | Active |
| 3 | A | Cancelled |
+-------------+--------------+----------------+
现在我正在尝试使用硬代码 table 进行交叉连接，根据我们投资组合中的所有 4 种产品以及我希望的状态，每个 customer_id 将给出 4 行喜欢申请
投资组合Table
+--------------+------------+----------+
| product_name | status_1 | status_2 |
+--------------+------------+----------+
| A | Inelegible | Inactive |
| B | Inelegible | Inactive |
| C | Ineligible | Inactive |
| D | Inelegible | Inactive |
+--------------+------------+----------+
在我的代码中，我尝试使用 CROSS JOIN 以实现每个 customer_id 4 行。不幸的是，对于拥有不止一种产品的客户，我有 double/triple 行。
这是我的代码：
SELECT
p.customer_id,
CASE WHEN p.product_name = pt.product_name THEN p.product_name ELSE pt.product_name END AS product_name,
CASE
WHEN p.product_name = pt.product_name THEN p.product_status
ELSE pt.status_1
END AS product_status
FROM
products AS p
CROSS JOIN
portfolio as pt
这是我当前的输出：
+----+-------------+--------------+----------------+
| # | customer_id | product_name | product_status |
+----+-------------+--------------+----------------+
| 1 | 1 | A | Active |
| 2 | 1 | B | Inelegible |
| 3 | 1 | C | Inelegible |
| 4 | 1 | D | Inelegible |
| 5 | 2 | A | Ineligible |
| 6 | 2 | A | Ineligible |
| 7 | 2 | B | Active |
| 8 | 2 | B | Ineligible |
| 9 | 2 | C | Active |
| 10 | 2 | C | Ineligible |
| 11 | 2 | D | Ineligible |
| 12 | 2 | D | Ineligible |
| 13 | 3 | A | Cancelled |
| 14 | 3 | B | Ineligible |
| 15 | 3 | C | Ineligible |
| 16 | 3 | D | Ineligible |
+----+-------------+--------------+----------------+
如您所见，对于 customer_id 2，我有两行用于每个产品，其中产品 B 和 C 的状态与我在 product_status table 上的状态不同.
在这种情况下，我想要实现的是一个具有 12 行的 table，其中当前 product/status 来自 product_status[=显示 44=] table，并添加投资组合 table 中剩余的 product/status。
预期输出
+----+-------------+--------------+----------------+
| # | customer_id | product_name | product_status |
+----+-------------+--------------+----------------+
| 1 | 1 | A | Active |
| 2 | 1 | B | Inelegible |
| 3 | 1 | C | Inelegible |
| 4 | 1 | D | Inelegible |
| 5 | 2 | A | Ineligible |
| 6 | 2 | B | Active |
| 7 | 2 | C | Active |
| 8 | 2 | D | Ineligible |
| 9 | 3 | A | Cancelled |
| 10 | 3 | B | Ineligible |
| 11 | 3 | C | Ineligible |
| 12 | 3 | D | Ineligible |
+----+-------------+--------------+----------------+
不确定 CROSS JOIN 是否是最佳选择，但现在我 运行 没主意了。
编辑：
我想到了另一个更清洁的解决方案。首先进行交叉连接，然后在 customer_id 和 product_name 上进行右连接，然后合并产品状态。
SELECT customer_id, product_name, coalesce(product_status, status_1)
FROM products p
RIGHT JOIN (
SELECT *
FROM (SELECT DISTINCT customer_id FROM products) pro
CROSS JOIN portfolio
) pt
USING (customer_id, product_name)
ORDER BY customer_id, product_name
旧答案：
这个想法是将customer_id的所有产品名称的信息包含到一个列表中，并检查组合中的产品是否在该列表中。
(SELECT customer_id, pt_product_name as product_name, first(status_1) as product_status
FROM (
SELECT
customer_id,
p.product_name as p_product_name,
pt.product_name as pt_product_name,
product_status,
status_1,
status_2,
collect_list(p.product_name) over (partition by customer_id) AS product_list
FROM products p
CROSS JOIN portfolio pt
)
WHERE NOT array_contains(product_list, pt_product_name)
GROUP BY customer_id, product_name)
UNION ALL
(SELECT customer_id, p_product_name as product_name, first(product_status) as product_status
FROM (
SELECT
customer_id,
p.product_name as p_product_name,
pt.product_name as pt_product_name,
product_status,
status_1,
status_2,
collect_list(p.product_name) over (partition by customer_id) AS product_list
FROM products p
CROSS JOIN portfolio pt)
WHERE array_contains(product_list, pt_product_name)
GROUP BY customer_id, product_name)
ORDER BY customer_id, product_name;
这给出了
+-----------+------------+--------------+
|customer_id|product_name|product_status|
+-----------+------------+--------------+
| 1| A| Active|
| 1| B| Inelegible|
| 1| C| Ineligible|
| 1| D| Inelegible|
| 2| A| Inelegible|
| 2| B| Active|
| 2| C| Active|
| 2| D| Inelegible|
| 3| A| Cancelled|
| 3| B| Inelegible|
| 3| C| Ineligible|
| 3| D| Inelegible|
+-----------+------------+--------------+
仅供参考 UNION ALL 之前的块给出：
+-----------+------------+--------------+
|customer_id|product_name|product_status|
+-----------+------------+--------------+
| 1| B| Inelegible|
| 1| C| Ineligible|
| 1| D| Inelegible|
| 2| A| Inelegible|
| 2| D| Inelegible|
| 3| B| Inelegible|
| 3| C| Ineligible|
| 3| D| Inelegible|
+-----------+------------+--------------+
UNION ALL 之后的块给出：
+-----------+------------+--------------+
|customer_id|product_name|product_status|
+-----------+------------+--------------+
| 1| A| Active|
| 2| B| Active|
| 2| C| Active|
| 3| A| Cancelled|
+-----------+------------+--------------+
希望对您有所帮助！
我正在尝试构建一个 table，它将结合我可以销售的所有产品，基于当前的产品。
产品状态Table
+-------------+--------------+----------------+
| customer_id | product_name | product_status |
+-------------+--------------+----------------+
| 1 | A | Active |
| 2 | B | Active |
| 2 | C | Active |
| 3 | A | Cancelled |
+-------------+--------------+----------------+
现在我正在尝试使用硬代码 table 进行交叉连接，根据我们投资组合中的所有 4 种产品以及我希望的状态，每个 customer_id 将给出 4 行喜欢申请
投资组合Table
+--------------+------------+----------+
| product_name | status_1 | status_2 |
+--------------+------------+----------+
| A | Inelegible | Inactive |
| B | Inelegible | Inactive |
| C | Ineligible | Inactive |
| D | Inelegible | Inactive |
+--------------+------------+----------+
在我的代码中，我尝试使用 CROSS JOIN 以实现每个 customer_id 4 行。不幸的是，对于拥有不止一种产品的客户，我有 double/triple 行。
这是我的代码：
SELECT
p.customer_id,
CASE WHEN p.product_name = pt.product_name THEN p.product_name ELSE pt.product_name END AS product_name,
CASE
WHEN p.product_name = pt.product_name THEN p.product_status
ELSE pt.status_1
END AS product_status
FROM
products AS p
CROSS JOIN
portfolio as pt
这是我当前的输出：
+----+-------------+--------------+----------------+
| # | customer_id | product_name | product_status |
+----+-------------+--------------+----------------+
| 1 | 1 | A | Active |
| 2 | 1 | B | Inelegible |
| 3 | 1 | C | Inelegible |
| 4 | 1 | D | Inelegible |
| 5 | 2 | A | Ineligible |
| 6 | 2 | A | Ineligible |
| 7 | 2 | B | Active |
| 8 | 2 | B | Ineligible |
| 9 | 2 | C | Active |
| 10 | 2 | C | Ineligible |
| 11 | 2 | D | Ineligible |
| 12 | 2 | D | Ineligible |
| 13 | 3 | A | Cancelled |
| 14 | 3 | B | Ineligible |
| 15 | 3 | C | Ineligible |
| 16 | 3 | D | Ineligible |
+----+-------------+--------------+----------------+
如您所见，对于 customer_id 2，我有两行用于每个产品，其中产品 B 和 C 的状态与我在 product_status table 上的状态不同.
在这种情况下，我想要实现的是一个具有 12 行的 table，其中当前 product/status 来自 product_status[=显示 44=] table，并添加投资组合 table 中剩余的 product/status。
预期输出
+----+-------------+--------------+----------------+
| # | customer_id | product_name | product_status |
+----+-------------+--------------+----------------+
| 1 | 1 | A | Active |
| 2 | 1 | B | Inelegible |
| 3 | 1 | C | Inelegible |
| 4 | 1 | D | Inelegible |
| 5 | 2 | A | Ineligible |
| 6 | 2 | B | Active |
| 7 | 2 | C | Active |
| 8 | 2 | D | Ineligible |
| 9 | 3 | A | Cancelled |
| 10 | 3 | B | Ineligible |
| 11 | 3 | C | Ineligible |
| 12 | 3 | D | Ineligible |
+----+-------------+--------------+----------------+
不确定 CROSS JOIN 是否是最佳选择，但现在我 运行 没主意了。
编辑：
我想到了另一个更清洁的解决方案。首先进行交叉连接，然后在 customer_id 和 product_name 上进行右连接，然后合并产品状态。
SELECT customer_id, product_name, coalesce(product_status, status_1)
FROM products p
RIGHT JOIN (
SELECT *
FROM (SELECT DISTINCT customer_id FROM products) pro
CROSS JOIN portfolio
) pt
USING (customer_id, product_name)
ORDER BY customer_id, product_name
旧答案： 这个想法是将customer_id的所有产品名称的信息包含到一个列表中，并检查组合中的产品是否在该列表中。
(SELECT customer_id, pt_product_name as product_name, first(status_1) as product_status
FROM (
SELECT
customer_id,
p.product_name as p_product_name,
pt.product_name as pt_product_name,
product_status,
status_1,
status_2,
collect_list(p.product_name) over (partition by customer_id) AS product_list
FROM products p
CROSS JOIN portfolio pt
)
WHERE NOT array_contains(product_list, pt_product_name)
GROUP BY customer_id, product_name)
UNION ALL
(SELECT customer_id, p_product_name as product_name, first(product_status) as product_status
FROM (
SELECT
customer_id,
p.product_name as p_product_name,
pt.product_name as pt_product_name,
product_status,
status_1,
status_2,
collect_list(p.product_name) over (partition by customer_id) AS product_list
FROM products p
CROSS JOIN portfolio pt)
WHERE array_contains(product_list, pt_product_name)
GROUP BY customer_id, product_name)
ORDER BY customer_id, product_name;
这给出了
+-----------+------------+--------------+
|customer_id|product_name|product_status|
+-----------+------------+--------------+
| 1| A| Active|
| 1| B| Inelegible|
| 1| C| Ineligible|
| 1| D| Inelegible|
| 2| A| Inelegible|
| 2| B| Active|
| 2| C| Active|
| 2| D| Inelegible|
| 3| A| Cancelled|
| 3| B| Inelegible|
| 3| C| Ineligible|
| 3| D| Inelegible|
+-----------+------------+--------------+
仅供参考 UNION ALL 之前的块给出：
+-----------+------------+--------------+
|customer_id|product_name|product_status|
+-----------+------------+--------------+
| 1| B| Inelegible|
| 1| C| Ineligible|
| 1| D| Inelegible|
| 2| A| Inelegible|
| 2| D| Inelegible|
| 3| B| Inelegible|
| 3| C| Ineligible|
| 3| D| Inelegible|
+-----------+------------+--------------+
UNION ALL 之后的块给出：
+-----------+------------+--------------+
|customer_id|product_name|product_status|
+-----------+------------+--------------+
| 1| A| Active|
| 2| B| Active|
| 2| C| Active|
| 3| A| Cancelled|
+-----------+------------+--------------+
希望对您有所帮助！