我继承了Animal类和Dog和Cat类。类动物具有属性X。我想为没有"X“属性的”狗“和具有"X”属性的“猫”生成XML。XmlIgnore在这里不能以我预期的方式工作。
我试图使用虚拟属性,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; }
}发布于 2019-04-25 22:52:44
即使你不再需要它了,我还是找到了这个问题的解决方案。
您可以为类的某些字段创建XmlAttributeOverrides并设置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,您也可以通过将attrs.XmlIgnore设置为false来执行相反的操作。
有关更多信息,请查看this。
https://stackoverflow.com/questions/55614252
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