47

是否可以将未知 XML 反序列化为如下对象?

 var xml = @"<Students><Student><Name>Arul</Name><Mark>90</Mark></Student></Students>";

 var serializer = new XmlSerializer(typeof(DynamicObject));

 dynamic students = serializer.Deserialize(new XmlTextReader(new StringReader(xml)));
4

1 回答 1

78

你可能想试试这个。

string xml = @"<Students>
                <Student ID=""100"">
                    <Name>Arul</Name>
                    <Mark>90</Mark>
                </Student>
                <Student>
                    <Name>Arul2</Name>
                    <Mark>80</Mark>
                </Student>
            </Students>";

dynamic students = DynamicXml.Parse(xml);

var id = students.Student[0].ID;
var name1 = students.Student[1].Name;

foreach(var std in students.Student)
{
    Console.WriteLine(std.Mark);
}

public class DynamicXml : DynamicObject
{
    XElement _root;
    private DynamicXml(XElement root)
    {
        _root = root;
    }

    public static DynamicXml Parse(string xmlString)
    {
        return new DynamicXml(XDocument.Parse(xmlString).Root);
    }

    public static DynamicXml Load(string filename)
    {
        return new DynamicXml(XDocument.Load(filename).Root);
    }

    public override bool TryGetMember(GetMemberBinder binder, out object result)
    {
        result = null;

        var att = _root.Attribute(binder.Name);
        if (att != null)
        {
            result = att.Value;
            return true;
        }

        var nodes = _root.Elements(binder.Name);
        if (nodes.Count() > 1)
        {
            result = nodes.Select(n => n.HasElements ? (object)new DynamicXml(n) : n.Value).ToList();
            return true;
        }

        var node = _root.Element(binder.Name);
        if (node != null)
        {
            result = node.HasElements || node.HasAttributes ? (object)new DynamicXml(node) : node.Value;
            return true;
        }

        return true;
    }
}

- 编辑 -

为了使它与 xml 命名空间一起工作,我添加了RemoveNamespaces方法。

public class DynamicXml : DynamicObject
{
    XElement _root;
    private DynamicXml(XElement root)
    {
        _root = root;
    }

    public static DynamicXml Parse(string xmlString)
    {
        return new DynamicXml(RemoveNamespaces(XDocument.Parse(xmlString).Root));
    }

    public static DynamicXml Load(string filename)
    {
        return new DynamicXml(RemoveNamespaces(XDocument.Load(filename).Root));
    }

    private static XElement RemoveNamespaces(XElement xElem)
    {
        var attrs = xElem.Attributes()
                    .Where(a => !a.IsNamespaceDeclaration)
                    .Select(a => new XAttribute(a.Name.LocalName, a.Value))
                    .ToList();

        if (!xElem.HasElements)
        {
            XElement xElement = new XElement(xElem.Name.LocalName, attrs);
            xElement.Value = xElem.Value;
            return xElement;
        }

        var newXElem = new XElement(xElem.Name.LocalName, xElem.Elements().Select(e => RemoveNamespaces(e)));
        newXElem.Add(attrs);
        return newXElem;
    }

    public override bool TryGetMember(GetMemberBinder binder, out object result)
    {
        result = null;

        var att = _root.Attribute(binder.Name);
        if (att != null)
        {
            result = att.Value;
            return true;
        }

        var nodes = _root.Elements(binder.Name);
        if (nodes.Count() > 1)
        {
            result = nodes.Select(n => n.HasElements ? (object)new DynamicXml(n) : n.Value).ToList();
            return true;
        }

        var node = _root.Element(binder.Name);
        if (node != null)
        {
            result = node.HasElements || node.HasAttributes ? (object)new DynamicXml(node) : node.Value;
            return true;
        }

        return true;
    }
}
于 2012-12-04T15:08:53.797 回答