SUM（）的多个值相关表

Question

我有一个MySQL数据库三个相关表：

• inventory_items定义仓库物品的种类
• inventory_transactions定义广告的添加或删除
数量
• 的位置上确定的位置说库存

Inventory_transactions有许多存储在给定inventory_item如何添加或删除一个quantity_offset从给定的位置。

我有以下查询，它将获得global_quantity（给定inventory_item_id的所有quantity_offsets的总和）以及location_quantity，只要知道位置ID，它就可以为特定位置执行相同的操作。

SELECT sku, name, SUM(quantity_offset) AS global_quantity, 
(
SELECT SUM(quantity_offset) 
    FROM inventory_transactions 
    WHERE inventory_items.inventory_item_id = inventory_transactions.inventory_item_id AND inventory_transactions.location_id = 1 
) AS location_quantity 
FROM inventory_items 
JOIN inventory_transactions ON inventory_items.inventory_item_id = inventory_transactions.inventory_item_id 
GROUP BY sku 

我想用被添加到每个位置的列结束了，所以结果可能有值，例如：

sku, name, global_quantity, denver_quantity, dallas_quantity, ft_wayne_quantity 

我的回退是要做到这一点对个人inventory_item页面（当我就知道库存物品，可以通过定位组），但能做到在一个单一的查询，并把它全部纳入一个表将是非常巨大的。

起初，我以为我可以通过迭代子查询，但无法设置列名（AS）编程。

非常感谢您提供任何帮助。

Answer 1

顺便说一句，而不是使用子查询SELECT列表中，你可以得到与上如果行是从给定的位置，有条件地返回quantity_offset的计算式的SUM聚合相同的结果。我也将使用左外连接到inventory_transactions表，这样我就一定要得到一个回行的每一行中inventory_items：

SELECT i.sku 
    , i.name 
    , SUM(t.quantity_offset) AS global_quantity 
    , SUM(IF(t.location_id = 1,t.quantity_offset,NULL)) AS location_1_quantity 
    FROM inventory_items i 
    LEFT 
    JOIN inventory_transactions t ON i.inventory_item_id = t.inventory_item_id 
    GROUP BY i.sku 

（这不回答你的问题，但我想让你意识到这一点）。

要回答你的问题，首先，SELECT语句无法动态地改变它在结果集中返回的列的数量或类型。你需要定义你想返回的每一列的表达式。 （在SQL SELECT statment文字的产生可动态完成，产生特定的SELECT语句，而是由SELECT语句返回的列由SELECT语句固定。）

如果我有一个相对固定的LOCATION_ID价值观，我需要的结果集指定，我会重复的条件表达式多次的总和，每进行一次位置或什么的，我想位置设置：

SELECT i.sku 
    , i.name 
    , SUM(t.quantity_offset) AS global_quantity 
    , SUM(IF(t.location_id = 1,t.quantity_offset,NULL)) AS location_1_quantity 
    , SUM(IF(t.location_id = 2,t.quantity_offset,NULL)) AS location_2_quantity 
    , SUM(IF(t.location_id = 3,t.quantity_offset,NULL)) AS location_3_quantity 
    , SUM(IF(t.location_id IN (1,2,3),t.quantity_offset,NULL)) AS location_123_quantity 
    , SUM(IF(t.location_id = 4,t.quantity_offset,NULL)) AS location_4_quantity 
    FROM inventory_items i 
    LEFT 
    JOIN inventory_transactions t ON i.inventory_item_id = t.inventory_item_id 
    GROUP BY i.sku 

如果要返回的列的集合需要真正动态，那么我不会将它们作为列返回。相反，我会将每个位置的SUM（quantity_offset）作为单独的行返回，然后在客户端处理转换为列式表示。

Answer 2

试试这个

SELECT sku, name, SUM(quantity_offset) AS global_quantity, 
     (
     SELECT SUM(quantity_offset) 
     FROM inventory_transactions 
      WHERE inventory_items.inventory_item_id = inventory_transactions.inventory_item_id AND inventory_transactions.location_id = 1 
      AND location = denver) AS denver_quantity , 
     (
     SELECT SUM(quantity_offset) 
     FROM inventory_transactions 
      WHERE inventory_items.inventory_item_id = inventory_transactions.inventory_item_id AND inventory_transactions.location_id = 1 
      AND location = dallas) AS dallas_quantity, 
     (
     SELECT SUM(quantity_offset) 
     FROM inventory_transactions 
      WHERE inventory_items.inventory_item_id = inventory_transactions.inventory_item_id AND inventory_transactions.location_id = 1 
      AND location = ft_wayne) AS ft_wayne_quantity 
    FROM inventory_items 
    JOIN inventory_transactions ON inventory_items.inventory_item_id = inventory_transactions.inventory_item_id 
    GROUP BY sku 

注意我用了location作为列名我不知道你有什么位置的列名，只是更换

Answer 3

您应该使用有条件的资金 - 这是最有效的方式（最快的查询）我知道的：

SELECT sku, name, SUM(quantity_offset) AS global_quantity, 
    sum(case when inventory_transactions.location_id = 1 then quantity_offset else 0 end) as loc1_quantitity, 
    sum(case when inventory_transactions.location_id = 2 then quantity_offset else 0 end) as loc2_quantitity, 
    sum(case when inventory_transactions.location_id = 3 then quantity_offset else 0 end) as loc3_quantitity 
FROM inventory_items JOIN inventory_transactions ON inventory_items.inventory_item_id = inventory_transactions.inventory_item_id 
GROUP BY sku