總和與GROUP BY的XQuery

Question

我嘗試轉換下面的SQL

SELECT SUM((((1-Discount)*Quantity)*[Unit Price])) AS AMOUNT 
FROM Orders 
LEFT JOIN [Order Details] 
    ON Orders.[Order ID] = [Order Details].[Order ID] 
WHERE ((([Order Date]) BETWEEN #1/1/2006# AND #2/1/2006#)); 

的訂單和訂單明細有1一對多的關係（1是訂單）

我目前擁有的代碼顯示每個訂單詳細信息的全部價格，但是不會允許我將每個訂單詳細信息的總價值合併在一起以獲取完整價值。樣品Orders.xml

<OrderID>10248</OrderID> 
     <CustomerID>WILMK</CustomerID> 
     <EmployeeID>5</EmployeeID> 
     <OrderDate>1996-07-04T00:00:00</OrderDate> 
     <RequiredDate>1996-08-01T00:00:00</RequiredDate> 
     <ShippedDate>1996-07-16T00:00:00</ShippedDate> 
     <ShipVia>3</ShipVia> 
     <Freight>32.38</Freight> 
     <ShipName>Vins et alcools Chevalier</ShipName> 
     <ShipAddress>59 rue de l&apos;Abbaye</ShipAddress> 
     <ShipCity>Reims</ShipCity> 
     <ShipPostalCode>51100</ShipPostalCode> 
     <ShipCountry>France</ShipCountry> 
    </Orders> 
    <Orders> 
     <OrderID>10249</OrderID> 
     <CustomerID>TRADH</CustomerID> 
     <EmployeeID>6</EmployeeID> 
     <OrderDate>1996-07-05T00:00:00</OrderDate> 
     <RequiredDate>1996-08-16T00:00:00</RequiredDate> 
     <ShippedDate>1996-07-10T00:00:00</ShippedDate> 
     <ShipVia>1</ShipVia> 
     <Freight>11.61</Freight> 
     <ShipName>Toms Spezialitäten</ShipName> 
     <ShipAddress>Luisenstr. 48</ShipAddress> 
     <ShipCity>Münster</ShipCity> 
     <ShipPostalCode>44087</ShipPostalCode> 
     <ShipCountry>Germany</ShipCountry> 
    </Orders> 

打樣訂單的`的

for $orderdetails in doc("Order Details.xml")/dataroot/Order_x0020_Details 
let $order := doc("Orders.xml")/dataroot/Orders[OrderID = $orderdetails/OrderID] 

where xs:dateTime($order/OrderDate/text()) gt xs:dateTime("1996-04-06T00:00:00") and xs:dateTime($order/OrderDate/text()) lt xs:dateTime("1997-04-05T00:00:00") 

return 
<Invoice_Amounts> 
{ 
    $order/OrderID 
} 
<Amount> 
{ 
sum(((1 - xs:double($orderdetails/Discount)) * xs:double($orderdetails/UnitPrice)) * xs:double($orderdetails/Quantity)) 
} 
</Amount> 
</Invoice_Amounts> 

Details.xml

<OrderID>10248</OrderID> 
<ProductID>11</ProductID> 
<UnitPrice>14</UnitPrice> 
<Quantity>12</Quantity> 
<Discount>0</Discount> 
</Order_x0020_Details> 
<Order_x0020_Details> 
<OrderID>10248</OrderID> 
<ProductID>42</ProductID> 
<UnitPrice>9.8</UnitPrice> 
<Quantity>10</Quantity> 
<Discount>0</Discount> 
</Order_x0020_Details> 
<Order_x0020_Details> 
<OrderID>10248</OrderID> 
<ProductID>72</ProductID> 
<UnitPrice>34.8</UnitPrice> 
<Quantity>5</Quantity> 
<Discount>0</Discount> 
</Order_x0020_Details> 
<Order_x0020_Details> 
<OrderID>10249</OrderID> 
<ProductID>14</ProductID> 
<UnitPrice>18.6</UnitPrice> 
<Quantity>9</Quantity> 
<Discount>0</Discount> 
</Order_x0020_Details> 

Answer 1

所以 - 據我所知，你已經知道該怎麼做的加入，所以要重點關注的是總和操作。要做到這一點，最簡單的方法是使用一個循環設置一個變量，然後您可以使用：

let $doc := 
    <docroot> 
     <Order id="1"> 
      <Discount>0.05</Discount> 
      <Quantity>3</Quantity> 
      <UnitPrice>15</UnitPrice> 
     </Order> 
     <Order id="2"> 
      <Discount>0</Discount> 
      <Quantity>15</Quantity> 
      <UnitPrice>20</UnitPrice> 
     </Order> 
    </docroot> 

(: collect the orders relevant to your query; 
    in your "real" version you'll want to filter, 
    or do your join, or whatever :) 
let $orders := $doc/Order 

(: generate a sequence containing the price for 
    each such order :) 
let $order_costs := 
    for $order in $orders 
     return ((1.0 - xs:double($order/Discount)) 
        * xs:double($order/UnitPrice) 
        * xs:double($order/Quantity)) 

(: generate your final result :) 
return sum($order_costs) 

您還可以通過參數中嵌入的for循環sum()避免變量賦值：

return sum(for $order in $orders 
      return ((1.0 - xs:double($order/Discount)) 
        * xs:double($order/UnitPrice) 
        * xs:double($order/Quantity))