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在试图弄清楚如何将 .NET System.ServiceModel.Channels.Message 对象转换为 XmlDocument 时，我想我可以这样做：

Message message = Message.CreateMessage(messageVersion, "SOAPAction");

using(var messageBuffer = message.CreateBufferedCopy(int.MaxValue))
{
    var stream = new MemoryStream();
    using(var xmlWriter = XmlWriter.Create(stream))
    {
        var xmlDocument = new XmlDocument();

        messageBuffer.WriteMessage(stream);
        stream.Flush();

        stream.Position = 0;
        xmlDocument.Load(stream);
        stream.Close();

        Debug.Writeline(xmlDocument.OuterXml);
    }
}

但是，这会导致 xmlDocument.Load(stream) 出现错误：

"Data at the root level is invalid. Line 1, position 1."

我现在意识到我应该将 Message 对象的 WriteMessage 与这样的 XmlWriter 对象结合使用：

Message message = Message.CreateMessage(messageVersion, "SOAPAction");
using(var messageBuffer = message.CreateBufferedCopy(int.MaxValue))
{
    var stream = new MemoryStream();
    using(var xmlWriter = XmlWriter.Create(stream))
    {
        var xmlDocument = new XmlDocument();

        messageBuffer.CreateMessage().WriteMessage(xmlWriter);
        xmlWriter.Flush();
        stream.Flush();

        stream.Position = 0;
        xmlDocument.Load(stream);

        xmlWriter.Close();
        stream.Close();

        Debug.WriteLine(xmlDocument.OuterXml);
    }
}

我确定我错过了一些基本点，但这引出了我的问题 - MessageBuffer 对象上的 WriteMessage(Stream stream) 有什么用？

4

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