问题描述

例如，我想删除或更改以下属性属性或添加新属性.有可能吗?

For example I want to remove or change below property attributes or add a new one. Is it possible?

[XmlElement("bill_info")]
[XmlIgnore]
public BillInfo BillInfo
{
  get { return billInfo; }
  set { billInfo = value; }
}

推荐答案

(编辑 - 我误读了原来的问题)

(edit - I misread the original question)

您不能添加实际属性(它们被烧录到 IL 中)；但是，使用 XmlSerializer 您不必这样做 - 您可以在构造函数中为 XmlSerializer 提供其他属性.但是，如果这样做，您需要小心地缓存 XmlSerializer 实例，否则它将为每个实例创建一个额外的程序集，这有点泄漏.(如果您使用仅采用 Type 的简单构造函数，则不会这样做).查看 XmlAttributeOverrides.

You cannot add actual attributes (they are burned into the IL); however, with XmlSerializer you don't have to - you can supply additional attributes in the constructor to the XmlSerializer. You do, however, need to be a little careful to cache the XmlSerializer instance if you do this, as otherwise it will create an additional assembly per instance, which is a bit leaky. (it doesn't do this if you use the simple constructor that just takes a Type). Look at XmlAttributeOverrides.

举个例子:

using System;
using System.Xml.Serialization;
 public class Person
{
    static void Main()
    {
        XmlAttributeOverrides overrides = new XmlAttributeOverrides();
        XmlAttributes attribs = new XmlAttributes();
        attribs.XmlIgnore = false;
        attribs.XmlElements.Add(new XmlElementAttribute("personName"));
        overrides.Add(typeof(Person), "Name", attribs);

        XmlSerializer ser = new XmlSerializer(typeof(Person), overrides);
        Person person = new Person();
        person.Name = "Marc";
        ser.Serialize(Console.Out, person);
    }
    private string name;
    [XmlElement("name")]
    [XmlIgnore]
    public string Name { get { return name; } set { name = value; } }
}

请注意；如果 xml 属性只是说明性的，那么有一种 second 方法可以为与数据绑定相关的事物添加属性，方法是使用 TypeDescriptor.CreateProperty 和 ICustomTypeDescriptor 或 TypeDescriptionProvider.恐怕比 xml 的情况复杂得多 - 并且不适用于所有代码 - 只是使用组件模型的代码.

Note also; if the xml attributes were just illustrative, then there is a second way to add attributes for things related to data-binding, by using TypeDescriptor.CreateProperty and either ICustomTypeDescriptor or TypeDescriptionProvider. Much more complex than the xml case, I'm afraid - and doesn't work for all code - just code that uses the component-model.

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