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

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北恋
北恋 2021-01-03 19:41

Im trying to deserialize the response from this simple web service

Im using the following code:

WebRequest request = WebRequest.Create(\"http://inb37         


        
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  • 2021-01-03 19:57

    Declaring XmlSerializer as

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

    is enough.

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  • 2021-01-03 19:58

    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);
    }
    
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  • 2021-01-03 20:12

    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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