xmlns=''> was not expected. - There is an error in XML document (2, 2)

49,435

Solution 1

Declaring XmlSerializer as

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

is enough.

Solution 2

You want to deserialize the XML and treat it as a fragment.

There's a very straightforward workaround available here. I've modified it for your scenario:

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);
}

Solution 3

I had the same error with Deserialize "xml string with 2 namespaces declared" into object.

<?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; }
}

By adding field with [XmlAttribute] into ErrorNotification class deserialization works.

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);
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Updated on July 09, 2022

Comments

  • user1384603
    user1384603 almost 2 years

    Im trying to deserialize the response from this simple web service

    Im using the following code:

    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());
    
    • Trevor Hart
      Trevor Hart over 8 years
      What happened with me when I got this error was I was serializing many objects to the same file separately then trying to deserialize said xml file to a list, what fixed it was just deleting the xmlns portion of the xml file, creating a custom list, then serializing only that said list, then I deserialized the xml file to the custom list. I know this has nothing to do with you but I'm going to leave it here because that was the problem I ran into that made me look at this question in the first place so maybe it will help somebody else.
  • user1384603
    user1384603 over 11 years
    Thanks for your time. I found user L.B answer very simple and useful.
  • Joe
    Joe over 10 years
    FYI L.B is right, slight variation if there is a Parent node though. In the xml I'm de-serializing the parent node is <PlatformResponse> which was throwing the exception "<Platform xmlns="> was not expected". new XmlRootAttribute("response")); didn't clear the error, new XmlRootAttribute("PlatformResponse")); did. Vote up L.B.
  • Admin
    Admin about 7 years
    500 bounty on it's way to you mate. Thanks.
  • Denise Skidmore
    Denise Skidmore about 3 years
    Also note if your root attribute has a namespace on it, that you may need to massage that XmlRootAttribute a little. xmlRoot.ElementName = reader.Name.Split(':').Last(); xmlRoot.Namespace = reader.NamespaceURI;