使用多个JOINS.SUM()产生错误的值

Question

我在SQL查询中获取一些基本发票信息,并在同一查询中计算订单总额和付款总额.这是我到目前为止:

SELECT
    orders.billerID, 
    orders.invoiceDate, 
    orders.txnID, 
    orders.bName, 
    orders.bStreet1, 
    orders.bStreet2, 
    orders.bCity, 
    orders.bState, 
    orders.bZip, 
    orders.bCountry, 
    orders.sName, 
    orders.sStreet1, 
    orders.sStreet2, 
    orders.sCity, 
    orders.sState, 
    orders.sZip, 
    orders.sCountry, 
    orders.paymentType, 
    orders.invoiceNotes, 
    orders.pFee, 
    orders.shipping, 
    orders.tax, 
    orders.reasonCode, 
    orders.txnType, 
    orders.customerID, 
    customers.firstName AS firstName, 
    customers.lastName AS lastName, 
    customers.businessName AS businessName, 
    orderStatus.statusName AS orderStatus, 
    SUM((orderItems.itemPrice * orderItems.itemQuantity))
      + orders.shipping + orders.tax AS orderTotal, 
    SUM(payments.amount) AS totalPayments                       <-- this sum
FROM
    orders 
    LEFT JOIN customers ON orders.customerID = customers.id 
    LEFT JOIN orderStatus ON orders.orderStatus = orderStatus.id
    LEFT JOIN payments ON payments.orderID = orders.id          <-- this join
    LEFT JOIN orderItems ON orderItems.orderID = orders.id 

除了totalPayments列之外,一切都来自查询.数据库中有一个付款,其值为(10.00).查询提供的值是20.00(正好是两倍).我的理论是,由于某种原因,查询是两次"汇总"付款金额列.任何人都可以为我阐明这一点吗？

谢谢您的帮助!

Answer 1

如果您在没有a的情况下运行查询group by,您会看到某些付款有多行.那是因为你也加入了订单商品.对于orderitem和payment的每个组合,结果集将包含一行.

一种解决方案是将总和更改为:

,    <earlier columns>    
,    (   select SUM(payments.amount) 
         from payments 
         where payments.orderID = orders.id
     ) AS totalPayments
,    <later columns>

这将确保多个订单项的付款不会多次汇总.