使用XmlSerializer保存列表时出现'XamlParseException'

Question

我尝试将自定义对象的两个列表保存为List<Vechicle>类型的第一个列表。

XmlSerializer SerializerObjVechicle = new XmlSerializer(typeof(List<Vechicle>));

然后我收到错误

  

＆＃34;未处理的类型异常   ＆＃39; System.Windows.Markup.XamlParseException＆＃39;发生在   PresentationFramework.dll＆＃34;

这是我的车辆类

[Serializable]
public class Vechicle
{

    private int _Id;
    private String _Registration;
    public Vechicle(int id,String registration)
    {
        Id = id;
        Registration = registration;
    }
    public override string ToString()
    {
        return Id.ToString() + " " + Registration;
    }

    #region getters/setters
    public int Id{
        get { return _Id; }
        set { _Id = value; }
    }

    public String Registration
    {
        get { return _Registration; }
        set { _Registration = value; }
    }

    #endregion
}

}

Answer 1

需要添加不带参数的构造函数。 或者从ISerializable接口继承类。

http://msdn.microsoft.com/en-us/library/vstudio/ms233843(v=vs.110).aspx

Answer 2

bogza.anton的回答是对的，你需要提供一个没有参数的构造函数，我给出一个这样的样本：

[Serializable]
public class Vechicle
{

    private int _Id;
    private String _Registration;
    public Vechicle()
    {
        _Id = 1;
        _Registration = "default name";
    }
    public Vechicle(int id, String registration)
    {
        Id = id;
        Registration = registration;
    }
    public override string ToString()
    {
        return Id.ToString() + " " + Registration;
    }

    #region getters/setters
    public int Id
    {
        get { return _Id; }
        set { _Id = value; }
    }

    public String Registration
    {
        get { return _Registration; }
        set { _Registration = value; }
    }

    #endregion
}

private void button1_Click(object sender, EventArgs e)
    {
        List<Vechicle> vList = new List<Vechicle>()
        {
            new Vechicle(),
            new Vechicle(),
            new Vechicle{Id=2, Registration="hello"},
            new Vechicle{Id = 100, Registration="world"}
        };

        XmlSerializer SerializerObjVechicle = new XmlSerializer(vList.GetType());
        FileStream fs = new FileStream("d:\\test.xml", FileMode.OpenOrCreate);
        SerializerObjVechicle.Serialize(fs, vList);
    }

我的测试结果如下：