How to get xml in a required format when serializing an object using XML serialization in .Net

Question

How do I get the attributes to the inner element rather than root element?

[XmlRoot("Root")]
public class Test
{
    string type=null;
    int value=0;
    public string Type
    {
        get { return type; }
        set { type=value; }
    }
    [XmlAttribute]
    public int Value
    {
        get { return type; }
        set { type=value; }
    }
}

will result into

<Root Value="">
    <Type>
    </Type>
</Root>

However I want

<Root >
    <Type Value="">
    </Type>
</Root>

Please help me out. Thanks in advance.

Benjol · Accepted Answer · 2009-10-21 06:04:28Z

2

XmlSerializer is intended to be a natural map between the object model and the xml; the answer, then, is to structure your DTO the same as your xml. In this case, by wrapping Test in a second object:

public class Root {
    public Test Type {get;set;}
}

The alternative is implementing IXmlSerializable, but that is effort, and easy to get wrong. It isn't uncommon to require a separate object representation for serialization purposes - it shouldn't be assumed that your "regular" business/data objects are necessarily directly suitable for serialization.

edited Oct 21, 2009 at 6:04

Benjol

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answered Oct 21, 2009 at 5:14

Marc Gravell

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Comments

bobbymcr · Accepted Answer · 2009-10-21 05:49:02Z

1

The XmlSerializer is not super customizable. The closest you can get to achieving this (without resorting to custom serialization) is a wrapper class:

[XmlRoot("Root")]
public class Test
{
    public TypeData Type { get; set; }

    // ...
}

class TypeData
{
    public string Data { get; set; }

    [XmlAttribute]
    public int Value { get; set; }
}

In that case, you'll end up with:

<Root>
  <Type Value="">
    <Data>...</Data>
  </Type>
</Root>

edited Oct 21, 2009 at 5:49

answered Oct 21, 2009 at 5:14

bobbymcr

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Comments

Arsen Mkrtchyan · Accepted Answer · 2009-10-21 06:19:25Z

0

[XmlRoot("Root")]
public class Test
{
    TypeClass type=null;
    [XmlElement("Type")]
    public TypeClass Type
    {
       get{return type;}
       set{type=value;}
    }
}
[XmlRoot("Type")]
public class TypeClass
{
   int _value=0;
   [XMlAttribute]
   public int Value
   {
      get{return _value;}
      set{_value=value;}
   }
}

try this

edited Oct 21, 2009 at 6:19

answered Oct 21, 2009 at 5:13

Arsen Mkrtchyan

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