在c#中使用linq to XML从返回null的XML中访问后代元素

Question

请人们帮助我，我需要消费一个从我的应用程序返回xml的web服务，下载xml的代码工作正常，但我需要从xml文件中提取值，但我不断从代码中获得null返回值，精确地GetLocationFromXml()方法是返回null的方法，GetLocationAsXMLFromHost()方法工作正常。

这是完整的类

using System;
using System.Collections.Generic;
using System.Linq;
using System.Text;
using AMIS.Core.DTOs;
using System.Net;
using System.Xml.Linq;
using System.Xml;
using System.Linq;
public class GeoLocationService
{
   private string _hostWebSite = "http://api.hostip.info";
    private readonly XNamespace _hostNameSpace = "http://www.hostip.info/api";
    private readonly XNamespace _hostGmlNameSpace = "http://www.opengis.net/gml";
public LocationInfo GetLocationInfoFromIPAddress(string userHostIpAddress)
{
    IPAddress ipAddress = null;
    IPAddress.TryParse(userHostIpAddress, out ipAddress);
    string xmlData = GetLocationAsXMLFromHost(ipAddress.ToString());
    LocationInfo locationInfo = GetLocationFromXml(xmlData);
    return locationInfo;
}
private string GetLocationAsXMLFromHost(string userHostIpAddress)
{
    WebClient webClient= new WebClient();
    string formattedUrl = string.Format(_hostWebSite + "/?ip={0}", userHostIpAddress);
    var xmlData = webClient.DownloadString(formattedUrl);
    return xmlData;
}
private LocationInfo GetLocationFromXml(string xmlData)
{
    LocationInfo locationInfo = new LocationInfo();
    var xmlResponse = XDocument.Parse(xmlData);
    var nameSpace = (XNamespace)_hostNameSpace;
    var gmlNameSpace = (XNamespace)_hostGmlNameSpace;
    try
    {
        locationInfo = (from x in xmlResponse.Descendants(nameSpace + "Hostip")
                        select new LocationInfo
                        {
                            CountryName = x.Element(nameSpace + "countryName").Value,
                            CountryAbbreviation = x.Element(nameSpace + "countryAbbrev").Value,
                            LocationInCountry = x.Element(gmlNameSpace + "name").Value
                        }).SingleOrDefault();
    }
    catch (Exception)
    {
        throw;
    }
    return locationInfo;
}
}

和XML文件在

下面

<?xml version="1.0" encoding="iso-8859-1"?>
<HostipLookupResultSet version="1.0.1" xmlns:gml="http://www.opengis.net/gml" xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance"  xsi:noNamespaceSchemaLocation="http://www.hostip.info/api/hostip-1.0.1.xsd">
<gml:description>This is the Hostip Lookup Service</gml:description>
<gml:name>hostip</gml:name>
<gml:boundedBy>
<gml:Null>inapplicable</gml:Null>
</gml:boundedBy>
<gml:featureMember>
<Hostip>
  <ip>41.78.8.3</ip>
  <gml:name>(Unknown city)</gml:name>
  <countryName>NIGERIA</countryName>
  <countryAbbrev>NG</countryAbbrev>
  <!-- Co-ordinates are unavailable -->
  </Hostip>
</gml:featureMember>
</HostipLookupResultSet>

Answer 1

考虑到这些注释，我怀疑问题可能很简单:

private string _hostNameSpace = "hostip.info/api";

应:

private string _hostNameSpace = "http://hostip.info/api";

(其他也一样)我个人会让XNamespace的值以:

开头

private static readonly XNamespace HostNameSpace = "http://hostip.info/api";

编辑:好吧，在摆弄了你的例子之后(它本来可以更短更完整)，我已经找出了问题所在:你正在寻找使用"主机名称空间"的元素-但是XML中的元素不在任何名称空间中。只要去掉那些命名空间位，它就可以正常工作了。