基于XmlChoiceIdentifier创建对象 [英] Create object based on XmlChoiceIdentifier

问题描述

我正在使用Activator（C＃）动态创建对象，这些类之一如下：

I am creating objects dynamically using Activator(C#) and one of these classes looks like:

class Driver
{
   Driver() { }

   [XmlChoiceIdentifier("ItemElementName")]
   [XmlElement("Bit16", typeof(DriverModule))]
   [XmlElement("Bit32", typeof(DriverModule))]
   [XmlElement("Bit64", typeof(DriverModule))]
   [XmlElement("Unified", typeof(DriverUnified))]
   public object Item { get; set; }
   [XmlIgnore]
   public ItemChoiceType ItemElementName { get; set; }

   // ... other serialization methods
}

当我使用激活器创建Driver类的实例时，得到以下对象：

When I create instance of Driver class using Activator I get following object:

obj.Item = null;
obj.ItemElementName = "Bit16"

ItemElementName默认设置，因为其枚举，但是如果基于此枚举设置Item，该如何设置？
再一次，我使用Activator动态创建了许多对象，因此我无法对其进行硬编码-是否有可能在类中获取此信息并正确创建Item属性？

ItemElementName is set by default, because its enum, but how to set Item if its based on this enum? Once again, I am creating many object dynamically with Activator, so I cant hardcode it - its possible to get this information in class and create Item property properly?

非常感谢！

推荐答案

ItemElementName 设置为 ItemChoiceType.Bit16 ，因为这是枚举中的第一项。因此，它的值是 0 ，但是您可以将其视为 Bit16 。通过激活器，您可以创建一个新实例。如果您不输入参数来设置属性，那么它们的值将是默认值。

ItemElementName is set to ItemChoiceType.Bit16 because that is the first item in the enumeration. Hence its value is 0 but you can see it as Bit16. By Activator you creates a new instance. If you don't put arguments in order to set your properties, then their values will be default ones.

我看到您那里有XmlChoiceIdentifier和其他XmlSerializer的东西。此属性的用途是：

I see that you have there XmlChoiceIdentifier and other XmlSerializer's stuff. The purpose of this attribute is to:

1. 请勿序列化 ItemElementName 属性。 / li>
   
2. 要根据 Item 的序列化值在反序列化后恢复 ItemElementName 。

1. Do not serialize ItemElementName property.
2. To restore ItemElementName after deserialization based on serialized value of Item.

这就是我可以根据给定的信息告诉您的信息...

That's what I can tell you basing on given information...

这里是一个使用XmlSerializer和XmlChoiceIdentifier的示例：

Here is an example that utilizes XmlSerializer along with XmlChoiceIdentifier:

public class Choices
{
    [XmlChoiceIdentifier("ItemType")]
    [XmlElement("Text", Type = typeof(string))]
    [XmlElement("Integer", Type = typeof(int))]
    [XmlElement("LongText", Type = typeof(string))]
    public object Choice { get; set; }

    [XmlIgnore]
    public ItemChoiceType ItemType;
}

[XmlType(IncludeInSchema = false)]
public enum ItemChoiceType
{
    Text,
    Integer,
    LongText
}

class Program
{
    static void Main(string[] args)
    {
        Choices c1 = new Choices();
        c1.Choice = "very long text"; // You can put here a value of String or Int32.
        c1.ItemType = ItemChoiceType.LongText; // Set the value so that its type match the Choice type (Text or LongText due to type of value is string).

        var serializer = new XmlSerializer(typeof(Choices));
        using (var stream = new FileStream("Choices.xml", FileMode.Create))
            serializer.Serialize(stream, c1);

        // Produced xml file.
        // Notice:
        // 1. LongText as element name
        // 2. Choice value inside the element
        // 3. ItemType value is not stored
        /*
        <?xml version="1.0"?>
        <Choices xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" xmlns:xsd="http://www.w3.org/2001/XMLSchema">
          <LongText>very long text</LongText>
        </Choices>
        */

        Choices c2;
        using (var stream = new FileStream("Choices.xml", FileMode.Open))
            c2 = (Choices)serializer.Deserialize(stream);

        // c2.ItemType is restored
    }
}

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