Logo Questions Linux Laravel Mysql Ubuntu Git Menu
 

C# XML Serializer won't store an attribute

This is my first question on Stack Overflow. Apologies in advance if I don't do things quite right while I'm learning how things work here.

Here is my code :

public void TestSerialize()
{
    ShoppingBag _shoppingBag = new ShoppingBag();
    Fruits _fruits = new Fruits();
    _fruits.testAttribute = "foo";

    Fruit[] fruit = new Fruit[2];
    fruit[0] = new Fruit("pineapple");
    fruit[1]= new Fruit("kiwi");

    _fruits.AddRange(fruit);

    _shoppingBag.Items = _fruits;

    Serialize<ShoppingBag>(_shoppingBag, @"C:\temp\shopping.xml");
}

public static void Serialize<T>(T objectToSerialize, string filePath) where T : class
{
    XmlSerializer serializer = new XmlSerializer(typeof(T));

    using (StreamWriter writer = new StreamWriter(filePath))
    {
        serializer.Serialize(writer, objectToSerialize);
    }
}

[Serializable]
public class ShoppingBag
{
    private Fruits _items;

    public Fruits Items
    {
        get { return _items; }
        set {_items = value; }
    }
}

public class Fruits : List<Fruit>
{
    public string testAttribute { get; set; }
}

[Serializable]
public class Fruit 
{
    public Fruit() { }

    public Fruit(string value)
    {
        Name = value;
    }

    [XmlAttribute("name")]
    public string Name { get; set; }
}

It produces this XML :

<?xml version="1.0" encoding="utf-8" ?> 
<ShoppingBag xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" xmlns:xsd="http://www.w3.org/2001/XMLSchema">
  <Items>
    <Fruit name="pineapple" /> 
    <Fruit name="kiwi" /> 
  </Items>
</ShoppingBag>

I don't understand why I am not getting <Items testAttribute="foo">

Please can anyone tell me what I need to add to my code so that the Serializer will write this attribute out?

Thanks,

like image 273
MartinC Avatar asked Nov 06 '22 09:11

MartinC


1 Answers

You need an intermediary class there:

class Program
{
    static void Main()
    {
        var shoppingBag = new ShoppingBag
        {
            Items = new ShoppingBagItems
            {
                Fruits = new List<Fruit>(new[] {
                    new Fruit { Name = "pineapple" },
                    new Fruit { Name = "kiwi" },
                }),
                TestAttribute = "foo"
            }
        };
        var serializer = new XmlSerializer(typeof(ShoppingBag));
        serializer.Serialize(Console.Out, shoppingBag);
    }
}

public class ShoppingBag
{
    public ShoppingBagItems Items { get; set; }
}

public class ShoppingBagItems
{
    [XmlElement("Fruit")]
    public List<Fruit> Fruits { get; set; }

    [XmlAttribute("testAttribute")]
    public string TestAttribute { get; set; }
}

public class Fruit
{
    [XmlAttribute("name")]
    public string Name { get; set; }
}

Also note that you don't need to decorate your classes with the [Serializable] attribute as it is used only for binary serialization. Another remark is that you don't need to derive from List<T>, simply use it as a property.

like image 118
Darin Dimitrov Avatar answered Nov 12 '22 09:11

Darin Dimitrov