Create xml rootNode via c#

Question

I want to create a xml document and root element like this:

<rdf:RDF xmlns:cim="http://iec.ch/TC57/2009/CIM-schema-cim14#"

xmlns:rdf="http://www.w3.org/1999/02/22-rdf-syntax-ns#">

I try create this like that:

 XmlDocument doc = new XmlDocument();
        XmlNode rootNode = doc.CreateElement("rdf:RDF xmlns:cim="http://iec.ch/TC57/2009/CIM-schema-cim14#" xmlns:rdf="http://www.w3.org/1999/02/22-rdf-syntax-ns#">");
        doc.AppendChild(rootNode);

        XmlNode userNode = doc.CreateElement("user");
        XmlAttribute attribute = doc.CreateAttribute("age");
        attribute.Value = "42";
        userNode.Attributes.Append(attribute);
        userNode.InnerText = "John Doe";
        rootNode.AppendChild(userNode);

        userNode = doc.CreateElement("user");
        attribute = doc.CreateAttribute("age");
        attribute.Value = "39";
        userNode.Attributes.Append(attribute);
        userNode.InnerText = "Jane Doe";
        rootNode.AppendChild(userNode);

        doc.Save("C:/xml-test.xml");

But i have exeption :The ' ' character, hexadecimal value 0x20, cannot be included in a name. Or so on.

How to make this element? Thanks.

Do you specifically want to build the XmlDocument element-by-element like that, or could you just use XmlDocument.LoadXml()? — Steve Wilkes
– Steve Wilkes, Commented Aug 9, 2013 at 7:37
I want to create XML document element by element from my List<MyObject> and from it properties; — user2545071
– user2545071, Commented Aug 9, 2013 at 7:39
You might also want to look into XmlSerialization which can take a list of objects and generate XML for them - also worth looking at xsd.exe that can take the appropriate XSD file, and generate C# (or VB.Net) classes for you that will serialise/deserialise via XmlSerialisation. — Rowland Shaw
– Rowland Shaw, Commented Aug 9, 2013 at 7:46

Rowland Shaw · Accepted Answer · 2013-08-09 07:44:13Z

3

The method you're using for building XML is actually building a tree of objects (rather than as the textual representation of them), for for Schemas, you have to tell the document about them:

XmlDocument doc = new XmlDocument();
XmlSchemaSet xss = new XmlSchemaSet();
xss.Add("cim", "http://iec.ch/TC57/2009/CIM-schema-cim14#");
xss.Add("rdf", "http://www.w3.org/1999/02/22-rdf-syntax-ns#");
doc.Schemas = xss;
XmlNode rootNode = doc.CreateElement("rdf:RDF"); // This overload assumes the document already knows about the rdf schema as it is in the Schemas set
doc.AppendChild(rootNode);

answered Aug 9, 2013 at 7:44

Rowland Shaw

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Mikael Östberg · Accepted Answer · 2013-08-09 08:00:27Z

If you can consider using Linq to XML, here's an alternative.

// Your data
var users = new List<User> {
    new User { Name = "John", Age = 42 },
    new User { Name = "Jane", Age = 39 }
};

// Project the data into XElements
var userElements = 
    from u in users     
    select 
        new XElement("user", u.Name, 
            new XAttribute("age", u.Age));

// Build the XML document, add namespaces and add the projected elements
var doc = new XDocument(
    new XElement("RDF",
        new XAttribute(XNamespace.Xmlns + "cim", 
            XNamespace.Get("http://iec.ch/TC57/2009/CIM-schema-cim14#")),
        new XAttribute(XNamespace.Xmlns + "rdf", 
            XNamespace.Get("http://www.w3.org/1999/02/22-rdf-syntax-ns#")),
        userElements
    )
);      

doc.Save(@"c:\xml-test.xml");

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