Question

反序列化响应

我使用以下代码：

WebRequest request = WebRequest.Create("http://inb374.jelastic.tsukaeru.net:8080/VodafoneDB/webresources/vodafone/04111111");    
WebResponse ws = request.GetResponse();
XmlSerializer s = new XmlSerializer(typeof(string));
string reponse = (string)s.Deserialize(ws.GetResponseStream());

Answer 1

将XmlSerializer声明为

XmlSerializer s = new XmlSerializer(typeof(string),new XmlRootAttribute("response"));

就够了。

Answer 2

您希望反序列化XML并将其视为片段。

有一个非常简单的解决方法here。我已根据您的情况对其进行了修改：

var webRequest = WebRequest.Create("http://inb374.jelastic.tsukaeru.net:8080/VodafoneDB/webresources/vodafone/04111111");

using (var webResponse = webRequest.GetResponse())
using (var responseStream = webResponse.GetResponseStream())
{
    var rootAttribute = new XmlRootAttribute();
    rootAttribute.ElementName = "response";
    rootAttribute.IsNullable = true;

    var xmlSerializer = new XmlSerializer(typeof (string), rootAttribute);
    var response = (string) xmlSerializer.Deserialize(responseStream);
}

Answer 3

我在将“已声明2个命名空间的xml字符串的XML字符串”反序列化为对象时遇到了相同的错误。

<?xml version="1.0" encoding="utf-8"?>
<vcs-device:errorNotification xmlns:vcs-pos="http://abc" xmlns:vcs-device="http://def">
    <errorText>Can't get PAN</errorText>
</vcs-device:errorNotification>

[XmlRoot(ElementName = "errorNotification", Namespace = "http://def")]
public class ErrorNotification
{
    [XmlAttribute(AttributeName = "vcs-pos", Namespace = "http://www.w3.org/2000/xmlns/")]
    public string VcsPosNamespace { get; set; }

    [XmlAttribute(AttributeName = "vcs-device", Namespace = "http://www.w3.org/2000/xmlns/")]
    public string VcsDeviceNamespace { get; set; }

    [XmlElement(ElementName = "errorText", Namespace = "")]
    public string ErrorText { get; set; }
}

通过将具有 [XmlAttribute] 的字段添加到ErrorNotification类反序列化中。

public static T Deserialize<T>(string xml)
{
    var serializer = new XmlSerializer(typeof(T));
    using (TextReader reader = new StringReader(xml))
    {
        return (T)serializer.Deserialize(reader);
    }
}

var obj = Deserialize<ErrorNotification>(xml);

的xmlns = '' ＆GT;没想到。 - XML文档中存在错误（2,2）

3 个答案: