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c#serializationxml-serialization
Muhammet Can TONBUL
Muhammet Can TONBUL

Reputation: 3538

How do I change the XmlElement name dynamically?

[XmlRoot("A")]
public class Foos
{
   [XmlElement("A1")]
   public List<Foo> FooList{ get; set; }
}


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

This code working as well. But its not dynamic. I want [XmlRoot("A")] to [XmlRoot(ConfigurationManager.AppSettings[someValue])]. But is throw to syntax error. Then i try this

    public class Foos
    {
       [XmlElement("A1")]
       public List<Foo> FooList{ get; set; }
    }
var serializer = new XmlSerializer(typeof(Foos),new XmlRootAttribute(ConfigurationManager.AppSettings[someValue]));

This is work just root element.I'm working. I could not change the "XmlElement" value of the FooList dynamically.There can be more than one element in the class. How do I change the XmlElement value of all of them dynamically?

Upvotes: 2

Views: 5949

Answers (1)

Mutlu Kaya
Mutlu Kaya

Reputation: 129

You need to use XmlAttributesOverrides in a right way. Please Check.

Working version of your code is here. 

public class Foos
{       
    public List<Foo> FooList { get; set; }
}


public class Foo
{
    public string Name { get; set; }
}

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

        var xmlAttributeForFoos = new XmlAttributes { XmlRoot = new XmlRootAttribute(ConfigurationManager.AppSettings["someFoosValue"]), XmlType = new XmlTypeAttribute(ConfigurationManager.AppSettings["someFoosValue"]) };
        var xmlAttributeForFooList = new XmlAttributes();
        var xmlAttributeForFoo = new XmlAttributes();

        xmlAttributeForFooList.XmlElements.Add(new XmlElementAttribute(ConfigurationManager.AppSettings["someFooValue"]));
        xmlAttributeForFoo.XmlElements.Add(new XmlElementAttribute(ConfigurationManager.AppSettings["someFooNameValue"]));

         var overrides = new XmlAttributeOverrides();
        overrides.Add(typeof(Foos), xmlAttributeForFoos);
        overrides.Add(typeof(Foos), "FooList", xmlAttributeForFooList);
        overrides.Add(typeof(Foo), "Name", xmlAttributeForFoo);

        XmlSerializer serializer = new XmlSerializer(typeof(Foos), overrides);

        var foos = new Foos
        {
            FooList = new List<Foo>
            {
                new Foo{Name = "FooName"}
            }
        };

        using (var stream = File.Open("file.xml", FileMode.OpenOrCreate))
        {
            serializer.Serialize(stream, foos);
        }
    }
}

App settings 

<appSettings>
  <add key="someFoosValue" value="SomeFoos"/>    
  <add key="someFooValue" value="SomeFoo"/>
  <add key="someFooNameValue" value="FooName"/>
</appSettings>

output

<SomeFoos xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" xmlns:xsd="http://www.w3.org/2001/XMLSchema">
  <SomeFoo>
    <FooName>FooName</FooName>
  </SomeFoo>
</SomeFoos>

Upvotes: 1

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