问题
In order to return useful information in SoapException.Detail
for an asmx web service, I took an idea from WCF and created a fault class to contain said useful information. That fault object is then serialised to the required XmlNode
of a thrown SoapException
.
I'm wondering whether I have the best code to create the XmlDocument
- here is my take on it:
var xmlDocument = new XmlDocument();
var serializer = new XmlSerializer(typeof(T));
using (var stream = new MemoryStream())
{
serializer.Serialize(stream, theObjectContainingUsefulInformation);
stream.Flush();
stream.Seek(0, SeekOrigin.Begin);
xmlDocument.Load(stream);
}
Is there a better way of doing this?
UPDATE: I actually ended up doing the following, because unless you wrap the XML in a <detail>
xml element, you get a SoapHeaderException
at the client end:
var serialiseToDocument = new XmlDocument();
var serializer = new XmlSerializer(typeof(T));
using (var stream = new MemoryStream())
{
serializer.Serialize(stream, e.ExceptionContext);
stream.Flush();
stream.Seek(0, SeekOrigin.Begin);
serialiseToDocument.Load(stream);
}
// Remove the xml declaration
serialiseToDocument.RemoveChild(serialiseToDocument.FirstChild);
// Memorise the node we want
var serialisedNode = serialiseToDocument.FirstChild;
// and wrap it in a <detail> element
var rootNode = serialiseToDocument.CreateNode(XmlNodeType.Element, "detail", "");
rootNode.AppendChild(serialisedNode);
UPDATE 2: Given John Saunders excellent answer, I've now started using the following:
private static void SerialiseFaultDetail()
{
var fault = new ServiceFault
{
Message = "Exception occurred",
ErrorCode = 1010
};
// Serialise to the XML document
var detailDocument = new XmlDocument();
var nav = detailDocument.CreateNavigator();
if (nav != null)
{
using (XmlWriter writer = nav.AppendChild())
{
var ser = new XmlSerializer(fault.GetType());
ser.Serialize(writer, fault);
}
}
// Memorise and remove the element we want
XmlNode infoNode = detailDocument.FirstChild;
detailDocument.RemoveChild(infoNode);
// Move into a root <detail> element
var rootNode = detailDocument.AppendChild(detailDocument.CreateNode(XmlNodeType.Element, "detail", ""));
rootNode.AppendChild(infoNode);
Console.WriteLine(detailDocument.OuterXml);
Console.ReadKey();
}
回答1:
EDIT: Creates output inside of detail element
public class MyFault
{
public int ErrorCode { get; set; }
public string ErrorMessage { get; set; }
}
public static XmlDocument SerializeFault()
{
var fault = new MyFault
{
ErrorCode = 1,
ErrorMessage = "This is an error"
};
var faultDocument = new XmlDocument();
var nav = faultDocument.CreateNavigator();
using (var writer = nav.AppendChild())
{
var ser = new XmlSerializer(fault.GetType());
ser.Serialize(writer, fault);
}
var detailDocument = new XmlDocument();
var detailElement = detailDocument.CreateElement(
"exc",
SoapException.DetailElementName.Name,
SoapException.DetailElementName.Namespace);
detailDocument.AppendChild(detailElement);
detailElement.AppendChild(
detailDocument.ImportNode(
faultDocument.DocumentElement, true));
return detailDocument;
}
来源:https://stackoverflow.com/questions/781442/serialize-object-to-xmldocument