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我有类 Animal 和类 Dog 和 Cat 继承自它。Animal 类具有属性 X。我想为没有“X”属性的“Dog”和具有“X”属性的“Cat”生成 XML。XmlIgnore 在这里没有按我预期的方式工作。

我尝试使用虚拟属性，然后在派生类中覆盖它，但它不起作用。

class Program
{
    static void Main(string[] args)
    {
        Dog dog = new Dog();
        Cat cat = new Cat();

        SerializeToFile(dog, "testDog.xml");
        SerializeToFile(cat, "testCat.xml");
    }

    private static void SerializeToFile(Animal animal, string outputFileName)
    {
        XmlSerializer serializer = new XmlSerializer(animal.GetType());

        TextWriter writer = new StreamWriter(outputFileName);
        serializer.Serialize(writer, animal);
        writer.Close();
    }
}

public abstract class Animal
{
    public virtual int X { get; set; }
}

public class Dog : Animal
{
    [XmlIgnore]
    public override int X { get; set; }
}

public class Cat : Animal
{
    public override int X { get; set; }
}
4

1 回答 1

0

即使您不再需要它，我仍然找到了解决此问题的方法。

您可以为类的某些字段创建XmlAttributeOverrides和设置XmlAttributes.XmlIgnore属性。

private static void SerializeToFile(Animal animal, string outputFileName)
{
    // call Method to get Serializer
    XmlSerializer serializer = CreateOverrider(animal.GetType()); 
    TextWriter writer = new StreamWriter(outputFileName);
    serializer.Serialize(writer, animal);
    writer.Close();
}

// Return an XmlSerializer used for overriding.
public XmlSerializer CreateOverrider(Type type)
{
    // Create the XmlAttributeOverrides and XmlAttributes objects.
    XmlAttributeOverrides xOver = new XmlAttributeOverrides();
    XmlAttributes attrs = new XmlAttributes();

    /* Setting XmlIgnore to true overrides the XmlIgnoreAttribute
     applied to the X field. Thus it won't be serialized.*/
    attrs.XmlIgnore = true;
    xOver.Add(typeof(Dog), "X", attrs);

    XmlSerializer xSer = new XmlSerializer(type, xOver);
    return xSer;
}

当然你也可以通过设置来做相反的attrs.XmlIgnore事情false。

查看此内容以获取更多信息。

于 2019-04-25T14:52:43.663 回答