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如何将枚举值序列化为int?

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Espo · 技术社区 · 16 年前

我想将枚举值序列化为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
}

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();

This answer (对于另一个问题)似乎表明枚举应该作为默认值序列化为int,但它似乎没有这样做。以下是我的输出:

<?xml version="1.0" encoding="utf-16"?>
<Request xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" xmlns:xsd="http://www.w3.org/2001/XMLSchema">
  <request>Confirmation</request>
</Request>

6 回复 | 直到 7 年前

149

miha 15 年前

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(假设父类是CustomClass),如下所示:

<CustomClass>
  <EnumValue>2</EnumValue>
</CustomClass>

Marc Gravell 16 年前

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

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

IXmlSerializable

Peter McG 16 年前

有关使用DataContractSerializer实现所需功能的有趣方法,请参阅下面的完整示例控制台应用程序:

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:

<Request xmlns="petermcg.wordpress.com" xmlns:i="http://www.w3.org/2001/XMLSchema-instance">
    <request>2</request>
</Request>

您需要对DataContractSerializer的System.Runtime.Serialization.dll程序集的引用。

Abhishek 7 年前

using System.Xml.Serialization;

public class Request
{    
    [XmlIgnore()]
    public RequestType request;

    public int RequestTypeValue
    {
      get 
      {
        return (int)request;
      } 
      set
      {
        request=(RequestType)value; 
      }
    }
}

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

Glenn 16 年前

看看System.Enum类。Parse方法将字符串或int表示形式转换为Enum对象,ToString方法将Enum对象转换为可以序列化的字符串。

Adriaan Davel 13 年前

由于您将显式非顺序值分配给枚举选项,因此我假设您希望能够一次指定多个值(二进制标志),那么接受的答案是您唯一的选项。传入PreBooking | PreBookingConfirmation的整数值为9,序列化程序将无法对其进行反序列化,但使用shim属性强制转换它会很好地工作。或者您可能只是错过了3个值:)