我将第二张桌子加入第一张桌子的总和不正确

Question

我有两个表：

1. 订单（编号，地点，订购日期，运费，amazon_fees，总金额，利润损失，加工费）
2. orderItemDetails（id，orderId，产品名称，产品费用）

我写了这个查询来获取数据，但没有得到想要的数据。因此，请更正我的查询，以便获得上图中的数据。

SELECT 
    o.totalshippingfees,
    o.totalamazonefees,
    o.totalorderamount,
    o.totalprofitloss,
    o.totalprocessing_fees,
    oi.totalproductexpense,
    o.site
FROM (
    SELECT 
        orderdate,
        site,
        sum(orders.shipping_fees) as totalshippingfees,
        sum(orders.amazone_fees) as totalamazonefees,
        sum(orders.totalAmount) as totalorderamount,
        sum(orders.profit_loss) as totalprofitloss,
        sum(orders.processing_fees) as totalprocessing_fees
    FROM orders
    group by site
) as o
JOIN (
    SELECT 
        site,
        sum(orderItemDetails.product_expense) as totalproductexpense,
        sum(orders.shipping_fees) as totalshippingfees,
        sum(orders.amazone_fees) as totalamazonefees,
        sum(orders.totalAmount) as totalorderamount,
        sum(orders.profit_loss) as totalprofitloss,
        sum(orders.processing_fees) as totalprocessing_fees
    from orders 
    LEFT JOIN orderItemDetails ON orders.id = orderItemDetails.orderId 
    group by site
) as oi
WHERE MONTH(o.orderdate) = '".$monthNo."' AND YEAR(o.orderdate)= '".$monthyear[1]."'

Answer 1

如果我理解正确，可以在site上使用orders进行简单分组，在site上使用orderItemDetails进行分组，然后在site上加入它们。

select *
from (
    select 
      sum(shipping_fees) as totalshippingfees,
      sum(amazone_fees) as totalamazonefees,
      sum(totalAmount) as totalorderamount,
      sum(profit_loss) as totalprofitloss,
      sum(processing_fees) as totalprocessing_fees,
      site
    from orders 
    group by site
 ) d1
 join (
        select
            sum(product_expense) as totalproductexpense,
            site
        from orders o
        left join orderItemDetails od
        on o.id = od.orderId 
        group by site
      ) d2
on d1.site = d2.site

DB小提琴： https://www.db-fiddle.com/f/dkpctq1XrzB7bfsrojaZHd/2