views:

73

answers:

1

Hi

I am trying to change the root name when doing XML serialization with C#.

It always takes the class name and not the name I am trying to set it with.

using System;
using System.Collections.Generic;
using System.Linq;
using System.Text;
using System.Xml.Serialization;
using System.IO;

namespace ConsoleApplication1
{
    class Program
    {
        static void Main(string[] args)
        {


            MyTest test = new MyTest();
            test.Test = "gog";

            List<MyTest> testList = new List<MyTest>() 
            {    
                test 
            }; 

            SerializeToXML(testList);
        }

        static public void SerializeToXML(List<MyTest> list)
        {
            XmlSerializer serializer = new XmlSerializer(typeof(List<MyTest>));
            TextWriter textWriter = new StreamWriter(@"C:\New folder\test.xml");
            serializer.Serialize(textWriter, list);
            textWriter.Close();
        }
    }


}





using System;
using System.Collections.Generic;
using System.Linq;
using System.Text;
using System.Xml.Serialization;

namespace ConsoleApplication1
{

    [XmlRootAttribute(ElementName = "WildAnimal", IsNullable = false)]
    public class MyTest
    {
        [XmlElement("Test")]
        public string Test { get; set; }


    }
}

Result

<?xml version="1.0" encoding="utf-8"?>
<ArrayOfMyTest xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" xmlns:xsd="http://www.w3.org/2001/XMLSchema"&gt;
  <MyTest>
    <Test>gog</Test>
  </MyTest>
</ArrayOfMyTest>

It does not change it to WildAnimal. I am not sure why. I got this from a tutorial.

Edit @ Marc

Thanks. I now see what your doing just seems so odd that you have to make a wrapper around it. I have one more question what happens if I wanted to make this format

<root>
   <element>
        <name></name>
   </element>
   <anotherElement>
       <product></product> 
   </anotherElement>
</root>

so like a nested elements. Would I have to make a new class for the second part and stick that in the wrapper class too?

A: 

In your exmaple, MyTest is not the root; are you trying to rename the array? I would write a wrapper:

[XmlRoot("NameOfRootElement")]
public class MyWrapper {
    private List<MyTest> items = new List<MyTest>();
    [XmlElement("NameOfChildElement")]
    public List<MyTest> Items { get { return items; } }
}

static void Main() {
    MyTest test = new MyTest();
    test.Test = "gog";

    MyWrapper wrapper = new MyWrapper {
        Items = {  test }
    };
    SerializeToXML(wrapper);
}

static public void SerializeToXML(MyWrapper list) {
    XmlSerializer serializer = new XmlSerializer(typeof(MyWrapper));
    using (TextWriter textWriter = new StreamWriter(@"test.xml")) {
        serializer.Serialize(textWriter, list);
        textWriter.Close();
    }
}
Marc Gravell
I am not sure what you mean by wrapper.
chobo2
@chobo2 - instead of serializing a `List<T>`... will make it a full example...
Marc Gravell
I guess I am trying to change the name of the "ArrayOfMyTest" I thought MyTest was the root and that was just something the serialization needed. So what do I do with this wrapper? Like how do I take it now and serialize it?
chobo2
@chobo2 - I've provided a full example with `Main`, `SerializeToXML` and `MyWrapper`.
Marc Gravell
See my edit....
chobo2