C# 获取深度嵌入的XML元素值

我正在尝试获取地址的纬度/经度，并使用dev.virtualearth.net上的XML提供程序

XML如下所示：

<Response xmlns:xsd="http://www.w3.org/2001/XMLSchema" 
    xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" 
    xmlns="http://schemas.microsoft.com/search/local/ws/rest/v1">
   <StatusCode>200</StatusCode>
   <StatusDescription>OK</StatusDescription>
   <AuthenticationResultCode>ValidCredentials</AuthenticationResultCode>
   <ResourceSets>
       <ResourceSet>
           <EstimatedTotal>2</EstimatedTotal>
       <Resources>
       <Location>
           <Name>350 Avenue V, New York, NY 11223</Name>
           <Point>
               <Latitude>40.595024898648262</Latitude>
               <Longitude>-73.969506248831749</Longitude>
           </Point>

但是对于纬度和经度值，我只是得到了null

---

我做错了吗？

您也应该将名称空间与嵌套元素一起使用

string xmlString = 
@"
<Response xmlns:xsd=""http://www.w3.org/2001/XMLSchema"" 
    xmlns:xsi=""http://www.w3.org/2001/XMLSchema-instance"" 
    xmlns=""http://schemas.microsoft.com/search/local/ws/rest/v1"">
   <StatusCode>200</StatusCode>
   <StatusDescription>OK</StatusDescription>
   <AuthenticationResultCode>ValidCredentials</AuthenticationResultCode>
   <ResourceSets>
       <ResourceSet>
           <EstimatedTotal>2</EstimatedTotal>
       <Resources>
       <Location>
           <Name>350 Avenue V, New York, NY 11223</Name>
           <Point>
               <Latitude>40.595024898648262</Latitude>
               <Longitude>-73.969506248831749</Longitude>
           </Point>
        </Location>
        </Resources>
        </ResourceSet>
    </ResourceSets>
</Response>
";
var doc = XDocument.Parse(xmlString);
XNamespace ns = "http://schemas.microsoft.com/search/local/ws/rest/v1";
var positions = doc.Descendants(ns + "Point")
       .Select(p =>
               new {
                      Latitude = (double)p.Element(ns + "Latitude"),
                      Longitude = (double)p.Element(ns + "Longitude")
                   });

string xmlString=
@"
200
好啊
有效存款
2.
纽约州纽约市第五大道350号，邮编：11223
40.595024898648262
-73.969506248831749
";
var doc=XDocument.Parse（xmlString）；
XNS=”http://schemas.microsoft.com/search/local/ws/rest/v1";
变量位置=单据子体（ns+“点”）
.选择（p=>
新的{
纬度=（双）p元素（ns+“纬度”），
经度=（双）p.Element（ns+“经度”）
});

我通常使用以下命令：var latlong=from c in doc.subjections（）。其中（x=>x.Name.LocalName==“Point”）

string xmlString = 
@"
<Response xmlns:xsd=""http://www.w3.org/2001/XMLSchema"" 
    xmlns:xsi=""http://www.w3.org/2001/XMLSchema-instance"" 
    xmlns=""http://schemas.microsoft.com/search/local/ws/rest/v1"">
   <StatusCode>200</StatusCode>
   <StatusDescription>OK</StatusDescription>
   <AuthenticationResultCode>ValidCredentials</AuthenticationResultCode>
   <ResourceSets>
       <ResourceSet>
           <EstimatedTotal>2</EstimatedTotal>
       <Resources>
       <Location>
           <Name>350 Avenue V, New York, NY 11223</Name>
           <Point>
               <Latitude>40.595024898648262</Latitude>
               <Longitude>-73.969506248831749</Longitude>
           </Point>
        </Location>
        </Resources>
        </ResourceSet>
    </ResourceSets>
</Response>
";
var doc = XDocument.Parse(xmlString);
XNamespace ns = "http://schemas.microsoft.com/search/local/ws/rest/v1";
var positions = doc.Descendants(ns + "Point")
       .Select(p =>
               new {
                      Latitude = (double)p.Element(ns + "Latitude"),
                      Longitude = (double)p.Element(ns + "Longitude")
                   });