Question

我在MySQL中遇到问题。我有2个表，customerorder和customerorderpos。 pos表包含已订购的项目 - 因此在customerorder表中每个订单总是只有一个条目，但是可能有几个在customerorderpos中具有相同的customerorderid。

除了我尝试对客户端进行求和计算外，查询中的所有内容都有效。例如sum（cart_total_complete）。这将返回重复值（对于具有多个customerorderpos记录的订单，它会多次计数）。

我确定它是基本的东西，无论是使用连接类型还是我如何使用不同的东西，但我已经尝试了几个小时而且无法让它发挥作用。 ..

任何想法我做错了什么？谢谢你的帮助!!

select concat(left(dayname(from_unixtime(customerorder.datetime)),3), ' ',
day(from_unixtime(customerorder.datetime))) as day, 
count(distinct customerorder.customerorderid) as count_totalorders, 
count(distinct customerorderpos.itemid) as count_differentitems, 
sum(customerorderpos.quantity_ordered - customerorderpos.quantity_cancelled) as quantity_ordered, 
sum(customerorderpos.itemsubtotal) as item_subtotal, 
sum(customerorderpos.pricechangetotal) as item_pricechangetotal, 
sum(customerorderpos.itemtotal) as item_total, 
sum(customerorderpos.purchase_price * (customerorderpos.quantity_ordered - customerorderpos.quantity_cancelled)) as item_purchasepricetotal, 
sum(cart_discounttotal) as total_discount, 
sum(customerorderpos.itemtotal) - sum(customerorderpos.purchase_price * (customerorderpos.quantity_ordered - customerorderpos.quantity_cancelled)) - sum(cart_discounttotal) as item_earningstotal, 
sum(cart_total_shipping) as total_shipping, 
sum(cart_total_tax) as total_tax, 
sum(cart_total_complete) as total_complete 
from customerorder inner join customerorderpos on customerorderpos.customerorderid = customerorder.customerorderid 
where customerorder.status_cancelled = 0 
group by day(from_unixtime(customerorder.datetime)) order by customerorder.datetime

Answer 1

如果您首先按customerorderpos汇总customerorderid表：

SELECT   customerorderid,
         SUM(quantity_ordered - quantity_cancelled) AS quantity_ordered,
         SUM(itemsubtotal) AS item_subtotal,
         SUM(pricechangetotal) AS item_pricechangetotal,
         SUM(itemtotal) AS item_total,
         SUM(purchase_price * (
           quantity_ordered - quantity_cancelled
         )) AS item_purchasepricetotal,
FROM     customerorderpos
GROUP BY customerorderid

然后，您可以将结果加入customerorder表：

SELECT   FROM_UNIXTIME(customerorder.datetime, '%a %e') AS day, 
         COUNT(*)                       AS count_totalorders,
         SUM(i.quantity_ordered)        AS quantity_ordered, 
         SUM(i.item_subtotal)           AS item_subtotal,
         SUM(i.item_pricechangetotal)   AS item_pricechangetotal, 
         SUM(i.item_total)              AS item_total, 
         SUM(i.item_purchasepricetotal) AS item_purchasepricetotal, 
         SUM(o.cart_discounttotal)      AS total_discount, 
         SUM(
           i.item_total
         - i.purchasepricetotal
         - o.cart_discounttotal
         )                              AS item_earningstotal, 
         SUM(o.cart_total_shipping)     AS total_shipping, 
         SUM(o.cart_total_tax)          AS total_tax, 
         SUM(o.cart_total_complete)     AS total_complete 
FROM     customerorder o JOIN (
  SELECT   customerorderid,
           COUNT(DISTINCT itemid) AS count_differentitems,
           SUM(quantity_ordered - quantity_cancelled) AS quantity_ordered,
           SUM(itemsubtotal) AS item_subtotal,
           SUM(pricechangetotal) AS item_pricechangetotal,
           SUM(itemtotal) AS item_total,
           SUM(purchase_price * (
             quantity_ordered - quantity_cancelled
           )) AS item_purchasepricetotal,
  FROM     customerorderpos
  GROUP BY customerorderid
) i USING (customerorderid)
WHERE    o.status_cancelled = 0 
GROUP BY FROM_UNIXTIME(o.datetime, '%e')
ORDER BY o.datetime

请注意，我已删除count_differentitems，因为不清楚是否要汇总每个订单中不同商品的数量（这将是扩展上述内容的简单案例）或者如果您希望所有聚合订单中的不同项目数（这需要额外的表连接）。

另请注意，我已使用FROM_UNIXTIME()的日期格式参数代替您手动尝试创建格式化日期字符串。

SUM计算MySQL连接中的重复记录

1 个答案: