MYSQL JOIN 和 SUM - 我做错了什么？ [关闭]

Question

我已经用尽了所有的选择。我不明白我哪里出了问题。请帮我。我不能接受更多的头撞。这就是我所在的地方...

下面是我的表结构的解释。

locations_management表
包含将locations_management_id、season_id、location_id和其他不需要的信息链接在一起的数据

订单表
包含有关订单的数据，包括 location_management_id 和 order_id 以及其他不需要的信息

orders_products 表
通过 order_id 链接到订单的产品数据。此表仅使用以下列：order_product_id、order_id、product_id、piece_qty

orders_adjustments表
用于跟踪对 inv_shipped 的任何调整。此表使用 order_id、product_id、piece_qty 列

这就是我今天所处的位置。下面的查询从上面的表中提取数据。

基本上，我要求从 locations_management 表 WHERE season_id = 12 AND location_id = 35 中获取 location_management_id。可以有多个可能的 location_management_id 同时适合 season_id 和 location_id。然后我需要找到与这些 location_management_id(s) 匹配的订单。找到订单后，我需要使用 order_id(s) 在 orders_products 表中查找与其关联的产品。 这个查询正是这样做的，但是当我进一步合并/求和时 piece_qty 为总 inv_shipped，数字发生了疯狂的事情。

SELECT  
    locations_management.season_id,  
    locations_management.location_id,  
    orders.order_id,  
    orders_products.product_id,  
    IFNULL((orders_products.piece_qty), 0) AS inv_shipped,  
    IFNULL((orders_adjustments.piece_qty), 0) AS inv_adjustments  
FROM  
    locations_management  
    JOIN orders USING (location_management_id)  
    LEFT JOIN orders_products USING (order_id)  
    LEFT JOIN orders_adjustments ON (orders_adjustments.order_id = orders_products.order_id) AND (orders_adjustments.product_id = orders_products.product_id)  
WHERE  
    locations_management.season_id = 12 AND locations_management.location_id = 35  
GROUP BY   
    product_id, orders_products.order_id

当我运行上面的查询时，这就是我得到的......

season_id   location_id     order_id    product_id      inv_shipped     inv_adjustments
12          35              2127        1               220             0
12          35              2194        1               160             0
12          35              2127        3               312             0
12          35              2127        4               24              0
12          35              2127        5               180             0
12          35              2194        5               24              0
12          35              2127        7               144             0
12          35              2127        7               24              0

这正是我期望得到的。多个 order_id 由 product_id 分组，所有数据都是准确的。所以现在这里成了问题。我想在 product_id 匹配并具有组合的 inv_shipped 时将它们添加/求和。所以 product_id 1 现在 inv_shipped 的总数为 380。

当我从上面获取相同的查询并将 SUM 添加到 inv_shipped 和 inv_adjustments（如下所示）时，我会在下面得到这个数据输出。注意一些值是如何翻倍的，但匹配的 product_id 行也没有合并。

    IFNULL(SUM(orders_products.piece_qty), 0) AS inv_shipped,  
    IFNULL(SUM(orders_adjustments.piece_qty), 0) AS inv_adjustments  

season_id   location_id     order_id    product_id      inv_shipped     inv_adjustments
12          35              2127        1               440             0
12          35              2194        1               160             0
12          35              2127        3               624             0
12          35              2127        4               48              0
12          35              2127        5               360             0
12          35              2194        5               24              0
12          35              2127        7               288             0
12          35              2127        7               24              0

如果我只将 GROUP BY 更改为 product_id，我会得到以下数据：

    GROUP BY product_id  

season_id   location_id     order_id    product_id      inv_shipped     inv_adjustments
12          35              2127        1               600             0
12          35              2127        3               624             0
12          35              2127        4               48              0
12          35              2127        5               384             0
12          35              2127        7               312             0

同样，这些 inv_shipped 总数不正确。那我哪里错了？

------------------------------------ 建议 ---------- --------------------------

建议使用以下查询，但 inv_shipped 的数据输出也未正确添加。

SELECT 
    locations_management.season_id,
    locations_management.location_id,
    orders.order_id,
    products.product_id,
    products.inv_shipped 
FROM
    locations_management
    JOIN (SELECT location_management_id, order_id FROM orders group by order_id) AS orders ON orders.location_management_id = locations_management.location_management_id
    JOIN (SELECT order_id, product_id, IFNULL(SUM(piece_qty), 0) AS inv_shipped FROM orders_products GROUP BY order_id, product_id) AS products ON products.order_id = orders.order_id

WHERE
    locations_management.season_id = 12 AND locations_management.location_id = 35
ORDER BY 
    product_id, order_id  
    
season_id   location_id     order_id    product_id      inv_shipped     inv_adjustments
12          35              2127        1               440             0
12          35              2194        1               160             0
12          35              2127        3               624             0
12          35              2127        4               48              0
12          35              2127        5               360             0
12          35              2194        5               24              0
12          35              2127        7               288             0
12          35              2127        7               24              0

Answer 1

首先，找到问题的根源。正确和错误信息之间有什么变化？

让我们看看您的第二个查询。据我所知，有三件事发生了变化：

1. 您已加入另一个子查询。
2. 您已添加SUM 操作。
3. 您已添加GROUP BY。

嵌套步骤是尝试删除SUM 和GROUP BY，如下所示：

SELECT 
    locations_management.season_id AS season_id,
    locations_management.location_id AS location_id,
    orders.order_id AS order_id,
    products.product_id AS product_id,
    products.piece_qty
FROM
    locations_management
    JOIN (SELECT location_management_id, order_id FROM orders group by order_id) AS orders ON orders.location_management_id = locations_management.location_management_id
    JOIN (SELECT order_id, product_id, piece_qty FROM orders_products) AS products ON products.order_id = orders.order_id

WHERE
    locations_management.season_id = 12 AND locations_management.location_id = 35

我假设每个product_id 将返回两（或更多）行。这可能是因为您的第二个JOIN 在orders_products 表中有两个（或更多）行order_id；这似乎很明显，因为表orders 的第一个子查询具有group by order_id。所以，现在要快速解决这个问题，您需要在第二个子查询中执行SUM。像这样的：

SELECT 
    locations_management.season_id AS season_id,
    locations_management.location_id AS location_id,
    orders.order_id AS order_id,
    products.product_id AS product_id,
    products.inv_shipped 
FROM
    locations_management
    JOIN (SELECT location_management_id, order_id FROM orders group by order_id) AS orders ON orders.location_management_id = locations_management.location_management_id
    JOIN (SELECT order_id, product_id, IFNULL(SUM(products.piece_qty), 0) AS inv_shipped FROM orders_products GROUP BY order_id, product_id) AS products ON products.order_id = orders.order_id

WHERE
    locations_management.season_id = 12 AND locations_management.location_id = 35;

这可能会返回正确的结果，但我个人会这样编写查询：

SELECT lm.season_id, lm.location_id, o.order_id , p.product_id, p.inv_shipped 
FROM locations_management AS lm
JOIN (SELECT location_management_id, order_id 
      FROM orders 
      GROUP BY location_management_id,order_id) AS o 
  ON o.location_management_id = lm.location_management_id
JOIN (SELECT order_id, product_id, IFNULL(SUM(products.piece_qty), 0) AS inv_shipped 
      FROM orders_products 
      GROUP BY order_id, product_id) AS p 
  ON p.order_id = o.order_id
WHERE
    lm.season_id = 12 AND lm.location_id = 35;

1. 如果您的别名与列名相同，则无需设置别名；例如lm.season_id AS season_id。如果您删除 .. AS season_id，该列仍将被识别为 season_id。您不会将其视为lm.season_id.. 至少对于我所知道的大多数工具而言。此外，我个人认为别名旨在缩短长表或列名，但 "to each their own"。
2. GROUP BY 应包括 SELECT 中的所有非聚合列。当然，如果sql_mode=only_full_group_by 关闭，您可以运行查询，但正确的设置应该是ON。你可以阅读更多of the reason why here。
3. 在GROUP BY 中添加额外的列后，此查询可能不会返回您曾经拥有的结果。这取决于您的数据，如果发生这种情况，我建议您edit your question 并添加Minimal, reproducible example。目前，我们只看到查询，没有可使用的示例表/数据。如果你能用几行数据create a fiddle就更好了。

Answer 2

我认为不需要那些嵌套的子查询。如果您希望每个订单和产品有一行，则可能不需要聚合。

你似乎想要这不是你想要的吗？

select lm.season_id, lm.location_id, 
    op.order_id, op.product_id, op.piece_qty as inv_shipped
from locations_management lm
inner join orders o on o.location_management_id = lm.location_management_id 
inner join order_products op on op.order_id = o.order_id
where lm.season_id = 12 and lm.location_id = 35

或者，如果您希望每个产品有一行：

select lm.season_id, lm.location_id, 
    op.product_id, coalesce(sum(op.piece_qty), 0) as inv_shipped
from locations_management lm
inner join orders o on o.location_management_id = lm.location_management_id 
left join order_products op on op.order_id = o.order_id
where lm.season_id = 12 and lm.location_id = 35
group by lm.season_id, lm.location_id, op.product_id