如何将枚举值序列化为int？

我想将我的枚举值序列化为int,但我只得到名称.

这是我的(示例)类和枚举:

public class Request {
    public RequestType request;
}

public enum RequestType
{
    Booking = 1,
    Confirmation = 2,
    PreBooking = 4,
    PreBookingConfirmation = 5,
    BookingStatus = 6
}

和代码(只是为了确保我做错了)

Request req = new Request();
req.request = RequestType.Confirmation;
XmlSerializer xml = new XmlSerializer(req.GetType());
StringWriter writer = new StringWriter();
xml.Serialize(writer, req);
textBox1.Text = writer.ToString();

这个答案(对另一个问题)似乎表明枚举应该序列化为默认值,但它似乎并没有这样做.这是我的输出:

  Confirmation

我已经能够通过在每个值上放置一个"[XmlEnum("X")]"属性来序列化为值,但这似乎是错误的.

1> miha..：

---

最简单的方法是使用[XmlEnum]属性,如下所示:

[Serializable]
public enum EnumToSerialize
{
    [XmlEnum("1")]
    One = 1,
    [XmlEnum("2")]
    Two = 2
}

这将序列化为XML(比如父类是CustomClass),如下所示:

  2

---

如果您的枚举使用[标志],这将不起作用

---

有争议的是,这实际上是否比接受的答案更少.更多枚举值=更多维护.

---

+1这也是我首选的方法 - 为什么在一些额外的指令允许你(de)正确序列化Enum时创建一个不必要的shim属性.

2> Marc Gravell..：

---

大多数时候,人们需要名字,而不是整数.您可以为此目的添加垫片属性吗？

[XmlIgnore]
public MyEnum Foo {get;set;}

[XmlElement("Foo")]
[EditorBrowsable(EditorBrowsableState.Never), Browsable(false)]
public int FooInt32 {
    get {return (int)Foo;}
    set {Foo = (MyEnum)value;}
}

或者你可以使用IXmlSerializable,但这是很多工作.

3> Peter McG..：

---

请参阅下面的完整示例控制台应用程序,以获得使用DataContractSerializer实现所需内容的有趣方法:

using System;
using System.IO;
using System.Runtime.Serialization;

namespace ConsoleApplication1
{
    [DataContract(Namespace="petermcg.wordpress.com")]
    public class Request
    {
        [DataMember(EmitDefaultValue = false)]
        public RequestType request;
    }

    [DataContract(Namespace = "petermcg.wordpress.com")]
    public enum RequestType
    {
        [EnumMember(Value = "1")]
        Booking = 1,
        [EnumMember(Value = "2")]
        Confirmation = 2,
        [EnumMember(Value = "4")]
        PreBooking = 4,
        [EnumMember(Value = "5")]
        PreBookingConfirmation = 5,
        [EnumMember(Value = "6")]
        BookingStatus = 6
    }

    class Program
    {
        static void Main(string[] args)
        {
            DataContractSerializer serializer = new DataContractSerializer(typeof(Request));

            // Create Request object
            Request req = new Request();
            req.request = RequestType.Confirmation;

            // Serialize to File
            using (FileStream fileStream = new FileStream("request.txt", FileMode.Create))
            {
                serializer.WriteObject(fileStream, req);
            }

            // Reset for testing
            req = null;

            // Deserialize from File
            using (FileStream fileStream = new FileStream("request.txt", FileMode.Open))
            {
                req = serializer.ReadObject(fileStream) as Request;
            }

            // Writes True
            Console.WriteLine(req.request == RequestType.Confirmation);
        }
    }
}

调用WriteObject后,request.txt的内容如下:

    2

您需要为DataContractSerializer引用System.Runtime.Serialization.dll程序集.