MySQL - 在选择中使用条件答案

【问题标题】：MySQL - Using condition from inside a selectMySQL - 在选择中使用条件
【发布时间】：2012-09-11 16:57:51
【问题描述】：

我有 3 张桌子：

订单：
- orders_id
- 购买日期
- ...
orders_products：
- orders_products_id
- orders_id
- products_id
- 产品名称
- products_quantity
- ...
products_warehouses：
- products_id
- warehouses_id
- product_quantity_in_warehouse

现在我想要一个包含以下信息的表格：

当前日期
products_id
产品名称
warehouses_id
current_quantity_in_warehouse
昨天售出

到目前为止，我得到了以下内容：

SELECT DISTINCT  CURRENT_DATE AS `date`
                 orpr.products_id AS products_id
                 orpr.products_name AS products_name
                 pwq.warehouses_id AS warehouses_id
                 pwq.products_quantity AS current_quantity_in_warehouse
                 (  SELECT SUM(orpr.products_quantity)
                    FROM `products_warehouses` wpq
                    INNER JOIN `orders_products` orpr ON pwq.products_id = orpr.products_id
                    INNER JOIN `orders` or ON orpr.orders_id = or.orders_id
                    WHERE pwq.products_id = ??
                    AND   pwq.warehouses_id = ?? ) AS sold_yeserday
FROM             `orders_products` orpr
INNER JOIN       `orders` or ON orpr.orders_id = or.orders_id
INNER JOIN       `products_warehouses pwq ON pwq.products_id = orpr.products_id;

我想要的是 ?? 的位置，即 products_id 和 warehouses_id 与结果的每一行类似。

如果有人有一个更简单的例子来说明如何在选择中使用选择（内部选择使用其中的选择条件 ID，当然这会改变每一行），它也很好，只要因为我可以重现该示例来解决我的问题。

提前致谢，

问候，凯文C

【问题讨论】：

这听起来像是一个相关的子查询，它有一些性能问题。将其转换为针对此聚合的连接可能更容易：JOIN (SELECT products_id, warehouses_id, SUM(products_quantity) GROUP BY products_id, warehouses_id)。
您能预览一下您预期输出的一些虚拟数据吗？如果可能的话，请提供您在sqlfiddle 上的当前进度示例，谢谢
ideone.com/5sb68 由于 SQL-fiddle 只允许 8000 个或更少的字符，我在 ideone.com 上发布了插入。由于隐私原因，我删除了所有不需要的数据，并替换了产品名称。（应将 16206 和 2 替换为该行的当前 ID）。

标签： mysql select parameters conditional-statements

【解决方案1】：

您应该提取内部 SELECT 语句并将结果放入表中，然后您可以在外部选择中加入该表

SELECT SUM(orpr.products_quantity) As 'Sold_yesterday', pwd.products_id, pwd.warehouses_id
INTO #tempTable
FROM `products_warehouses` wpq
INNER JOIN `orders_products` orpr ON pwq.products_id = orpr.products_id
INNER JOIN `orders` or ON orpr.orders_id = or.orders_id


SELECT DISTINCT  CURRENT_DATE AS `date`
orpr.products_id AS products_id
orpr.products_name AS products_name
pwq.warehouses_id AS warehouses_id
pwq.products_quantity AS current_quantity_in_warehouse                
FROM             `orders_products` orpr
INNER JOIN       `orders` or ON orpr.orders_id = or.orders_id
INNER JOIN       `products_warehouses pwq ON pwq.products_id = orpr.products_id;
INNER JOIN       '#tempTable' tt On tt.products_id = pwd.products_id AND tt.warehouses_id = pwd.warehouses_id

【讨论】：

我尝试了您的解决方案，但在#tempTable 部分出现错误。

【解决方案2】：

SELECT
    CURRENT_DATE AS `date`,
    orpr.products_id AS products_id,
    orpr.products_name AS products_name,
    pwq.warehouses_id AS warehouses_id,
    pwq.products_quantity AS current_quantity_in_warehouse,
    SUM(orpr.products_quantity) AS sold_yeserday
FROM             
    orders_products orpr
    INNER JOIN       
    products_warehouses pwq ON pwq.products_id = orpr.products_id;
group by orpr.products_id, pwq.warehouses_id
order by orpr.products_id, pwq.warehouses_id

【讨论】：