CODEWITHSUNDEEP
c#web-servicessoap
Michael G
Michael G

Reputation: 6745

XmlSerializer - Deserialize SoapException Detail Element

I'm trying to Deserialize a SoapException message that I am receiving from a web service. I'm getting the following exception:

System.InvalidOperationException was unhandled
  Message="There is an error in XML document (2, 2)."
  Source="System.Xml"
  StackTrace:
       at System.Xml.Serialization.XmlSerializer.Deserialize(XmlReader xmlReader, String encodingStyle, XmlDeserializationEvents events)
       at System.Xml.Serialization.XmlSerializer.Deserialize(XmlReader xmlReader)
       at FullServiceServices.AddressCorrectionService.GetResponse(FullServiceAddressCorrectionQueryRequest request) in AddressCorrectionService.cs:line 169
       at FullServiceServices.AddressCorrectionService.GenerateRequest(DateTime startDate, DateTime endDate) in AddressCorrectionService.cs:line 79
       at FullServiceServices.Form1.Form1_Load(Object sender, EventArgs e) in \Form1.cs:line 33

  InnerException: System.InvalidOperationException
       Message="<Fault xmlns='http://idealliance.org/maildat/Specs/md091/mailxml81/mailxml'> was not expected."
       Source="gn4ggxxc"
       StackTrace:
            at Microsoft.Xml.Serialization.GeneratedAssembly.XmlSerializationReaderFault.Read3_Fault()

How can I properly deserialize this XML into the provided partial class Fault?

Below is how I'm reading the XML string in to the XMLSerializer:

            XmlTextReader xr = new XmlTextReader(new StringReader(soapEx.Detail.FirstChild.InnerText));
            XmlSerializer xs = new XmlSerializer(typeof(Fault));
            Fault fault = xs.Deserialize(xr) as Fault;

Here is the InnerText of the soap exception .Detail property's FirstChild

<?xml version="1.0" encoding="UTF-8"?>
<mailxml:Fault xmlns:mailxml="http://idealliance.org/maildat/Specs/md091/mailxml81/mailxml" xmlns:mailxml_base="http://idealliance.org/maildat/Specs/md091/mailxml81/base" xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" xsi:schemaLocation="http://idealliance.org/maildat/Specs/md091/mailxml81/mailxml mailxml_031910.xsd http://idealliance.org/maildat/Specs/md091/mailxml81/base mailxml_base_031910.xsd">
  <mailxml:FaultCode>402</mailxml:FaultCode>
  <mailxml:FaultDescription>402 Not Well Formed XML</mailxml:FaultDescription>
</mailxml:Fault>

Here is the partial class that is auto-generated via WSDL.

/// <remarks/>
[System.CodeDom.Compiler.GeneratedCodeAttribute("System.Xml", "2.0.50727.3082")]
[System.SerializableAttribute()]
[System.Diagnostics.DebuggerStepThroughAttribute()]
[System.ComponentModel.DesignerCategoryAttribute("code")]
[System.Xml.Serialization.XmlTypeAttribute(AnonymousType=true, Namespace="http://idealliance.org/maildat/Specs/md091/mailxml81/mailxml")]
public partial class Fault {

    private string[] faultCodeField;

    private string[] faultDescriptionField;

    private string trackingIDField;

    /// <remarks/>
    [System.Xml.Serialization.XmlElementAttribute("FaultCode")]
    public string[] FaultCode {
        get {
            return this.faultCodeField;
        }
        set {
            this.faultCodeField = value;
        }
    }

    /// <remarks/>
    [System.Xml.Serialization.XmlElementAttribute("FaultDescription")]
    public string[] FaultDescription {
        get {
            return this.faultDescriptionField;
        }
        set {
            this.faultDescriptionField = value;
        }
    }

    /// <remarks/>
    [System.Xml.Serialization.XmlAttributeAttribute(Form=System.Xml.Schema.XmlSchemaForm.Qualified)]
    public string TrackingID {
        get {
            return this.trackingIDField;
        }
        set {
            this.trackingIDField = value;
        }
    }
}

Thanks!

Upvotes: 3

Views: 4903

Answers (2)

Michael G
Michael G

Reputation: 6745

Setting the namespace on the RootElement did the trick

XmlRootAttribute xRoot = new XmlRootAttribute();
xRoot.Namespace = "http://namespacehere";

XmlReader xr = soapEx.Detail.FirstChild.CreateNavigator().ReadSubtree();
XmlSerializer xs = new XmlSerializer(typeof(Fault), xRoot);

Fault fault = xs.Deserialize(xr) as Fault;

Upvotes: 4

John Saunders
John Saunders

Reputation: 161773

First, don't use new XmlTextReader() unless you have no other choices. It's been deprecated since .NET 2.0.

Second, you already have XML - why turn it into a string and then back into XML? Try

XmlReader xr = soapex.Detail.FirstChild.CreateNavigator().ReadSubTree();
Fault fault = xs.Deserialize(xr) as Fault;

Upvotes: 1

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