我无法使用XML序列化将动物列表保存到磁盘。
我收到Exception:Thrown:“不期望使用AnimalLibrary.Animals.Mammals.Dog类型。使用XmlInclude或SoapInclude属性指定静态未知的类型。” (System.InvalidOperationException)
如果我使用“Dog”尝试注释代码,它将按预期工作,并生成XML。但是作为列表中唯一元素发送的同一只狗不起作用。
[XmlElement("animalList")]
public List<Animal> animalList = new List<Animal>();
public bool SaveBinary(string fileName)
{
Mammals.Dog dog = (Mammals.Dog)animalList[0];
//IObjectSerializer<Mammals.Dog> obj = new XMLObjectSerializer<Mammals.Dog>();
IObjectSerializer<List<Animal>> obj = new XMLObjectSerializer<List<Animal>>();
bool saved = obj.SaveFile(fileName, animalList);
if (saved)
return true;
return false;
}
XML序列化程序
public bool SaveFile(string fileName, T objectToSerialize)
{
try
{
//Will overwrite old file
XmlSerializer mySerializer = new XmlSerializer(typeof(T));
StreamWriter myWriter = new StreamWriter(fileName);
mySerializer.Serialize(myWriter, objectToSerialize);
myWriter.Close();
}
catch (IOException ex)
{
Console.WriteLine("IO Exception ", ex.Message);
return false;
}
return true;
}
继承狗的文件。类中没有xml标签。
[XmlRoot(ElementName="Animal")]
public abstract class Animal : IAnimal
{
/// <summary>
/// Id of animal
/// </summary>
private string id;
public string ID
........
[XmlRoot(ElementName = "Animals")]
public abstract class Mammal : Animal
{
public int NumberofTeeth { get; set; }
........
[XmlRoot(ElementName="Dog")]
public class Dog : Mammal
{
/// <summary>
/// Constructor - Create an instance of a Dog
/// </summary>
public Dog()
{
}
........
答案 0 :(得分:12)
如果您想要一个对象列表并将它们序列化为基类型列表,那么您需要告诉序列化器可以使用哪种具体类型。
因此,如果您想将Dog和Cat对象放入Animal列表,则需要将标记添加到Animal类中,如下所示
[XmlInclude(typeof(Cat))]
[XmlInclude(typeof(Dog))]
[XmlRoot(ElementName="Animal")]
public abstract class Animal : IAnimal