MySQL 显示总和 0

Question

所以我有一个包含一组订单项目的表，订单项目表跟踪特定订单中包含的特定项目的数量。下面是本示例中使用的表格的简化版本。

订单商品表

order_item_id : int
order_item_item_id : int
order_item_quantity : int
order_item_order_id : int

订单表

order_id : int
order_date : date

我正在尝试执行单个查询，该查询将平均每周、每月和所有时间特定 order_item_item_id 的销售额。

目前我的查询如下

SELECT totalAvg.item 

ROUND(AVG(totalAvg.total)+0.4999999, 0) AS totalAverage,  
ROUND(AVG(monthAvg.total)+0.4999999, 0) AS monthAverage, 
ROUND(AVG(weekAvg.total)+0.4999999, 0) AS weekAverage from

(SELECT orderitem.order_item_item_id AS item, SUM(orderitem.order_item_quantity) AS total,
 DATE(`order`.order_date) AS date FROM orderitem, `order` WHERE orderitem.order_item_order_id = 
`order`.order_id GROUP BY DATE(`order`.order_date), order_item_item_id) AS totalAvg,

(SELECT orderitem.order_item_item_id AS item, SUM(orderitem.order_item_quantity) AS total, 
DATE(`order`.order_date) AS date FROM orderitem, `order` WHERE orderitem.order_item_order_id =
`order`.order_id AND `order`.order_date > date_sub(now(), interval 1 week) GROUP BY 
DATE(`order`.order_date), order_item_item_id) AS weekAvg,

(SELECT orderitem.order_item_item_id AS item, SUM(orderitem.order_item_quantity) AS total,
DATE(`order`.order_date) AS date FROM orderitem, `order` WHERE orderitem.order_item_order_id = 
`order`.order_id AND `order`.order_date > date_sub(now(), interval 1 month) GROUP BY 
DATE(`order`.order_date), order_item_item_id) AS monthAvg,

WHERE totalAvg.item = monthAvg.item AND
monthAvg.item = weekAvg.item 
GROUP BY item

这样做的问题是，如果 weekAvg 表中不存在某个项目，则不会为 totalAvg 或 monthAvg 打印任何结果。我怎样才能在单个语句中执行此操作？

样本数据

{order_id : 5, order_date : 02/02/2013}
{order_id : 6, order_date : 13/03/2013}

{order_item_id : 1, order_item_order_id : 5, order_item_item_id : 1, order_item_quantity : 3}
{order_item_id : 2, order_item_order_id : 6, order_item_item_id : 1, order_item_quantity : 4}

当前输出不会返回任何内容，因为每周报告没有 id 为 1 的订单项的条目。我正在尝试解决这个问题，以便它输出以下内容

totalAverage = 3.5
monthAverage = 4
weekAverage = 0

数值计算如下：

总平均数 = 特定商品的售出总数除以该商品的售出天数。

月和周值与之前的计算相同，但有时间限制，因此订单必须在上周/月内下达。

Answer 1

您需要使用outer joins 将totalAvg 连接到monthAvg 和weekAvg。您可以使用 COALESCE() 来确保您得到 0 而不是 NULL 来表示空的周或月值。

您还应该使用 ANSI 连接语法。它更不容易出错并且更易读。

试试这样的：

SELECT totalAvg.item,
ROUND(AVG(totalAvg.total)+0.4999999, 0) AS totalAverage,  
COALESCE(ROUND(AVG(monthAvg.total)+0.4999999, 0),0) AS monthAverage, 
COALESCE(ROUND(AVG(weekAvg.total)+0.4999999, 0),) AS weekAverage 
from
(
  SELECT orderitem.order_item_item_id AS item, 
    SUM(orderitem.order_item_quantity) AS total,
    DATE(`order`.order_date) AS date 
  FROM orderitem
    inner join `order` on orderitem.order_item_order_id = `order`.order_id 
  GROUP BY DATE(`order`.order_date), order_item_item_id
) AS totalAvg
left outer join
(
  SELECT orderitem.order_item_item_id AS item, 
    SUM(orderitem.order_item_quantity) AS total, 
    DATE(`order`.order_date) AS date 
  FROM orderitem
    inner join `order` on orderitem.order_item_order_id = `order`.order_id 
  WHERE `order`.order_date > date_sub(now(), interval 1 week) 
  GROUP BY 
  DATE(`order`.order_date), order_item_item_id
) AS weekAvg on weekAvg.item = totalAvg.item
left outer join
(
  SELECT orderitem.order_item_item_id AS item, 
    SUM(orderitem.order_item_quantity) AS total,
    DATE(`order`.order_date) AS date 
  FROM orderitem
    inner join `order` on orderitem.order_item_order_id = `order`.order_id 
  WHERE `order`.order_date > date_sub(now(), interval 1 month) 
  GROUP BY DATE(`order`.order_date), order_item_item_id
) AS monthAvg on monthAvg.item = totalAvg.item
GROUP BY item