将对象序列化为字符串

Question

我有以下方法将对象保存到文件:

// Save an object out to the disk
public static void SerializeObject<T>(this T toSerialize, String filename)
{
    XmlSerializer xmlSerializer = new XmlSerializer(toSerialize.GetType());
    TextWriter textWriter = new StreamWriter(filename);

    xmlSerializer.Serialize(textWriter, toSerialize);
    textWriter.Close();
}

我承认我没有写它(我只将其转换为带有类型参数的扩展方法).

现在我需要它将xml作为字符串返回给我(而不是将其保存到文件中).我正在研究它,但我还没想到它.

我认为对于熟悉这些对象的人来说这可能很容易.如果不是,我最终会弄明白.

Answer 1

使用StringWriter而不是StreamWriter:

public static string SerializeObject<T>(this T toSerialize)
{
    XmlSerializer xmlSerializer = new XmlSerializer(toSerialize.GetType());

    using(StringWriter textWriter = new StringWriter())
    {
        xmlSerializer.Serialize(textWriter, toSerialize);
        return textWriter.ToString();
    }
}

注意,使用toSerialize.GetType()而不是typeof(T)在XmlSerializer构造函数中使用是很重要的:如果使用第一个代码,代码将涵盖所有可能的子类T(对方法有效),而使用后者则在传递派生类型时会失败T.下面是一些示例代码的链接,它激发了这个语句,XmlSerializer抛出了一个Exceptionwhen typeof(T),因为你将一个派生类型的实例传递给一个方法,该方法调用派生类型的基类中定义的SerializeObject:http:// ideone .com/1Z5J1.

此外,Ideone使用Mono执行代码; Exception使用Microsoft .NET运行时的实际Message情况与Ideone上显示的实际情况有所不同,但它的失败方式却相同.

Answer 2

我知道这不是问题的答案,但根据问题的投票数和接受的答案,我怀疑人们实际上是在使用代码将对象序列化为字符串.

使用XML序列化会在输出中添加不必要的额外文本垃圾.

对于以下课程

public class UserData
{
    public int UserId { get; set; }
}

它会产生

<?xml version="1.0" encoding="utf-16"?>
<UserData xmlns:xsd="http://www.w3.org/2001/XMLSchema" 
          xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance">
   <UserId>0</UserId>
</UserData>

更好的解决方案是使用JSON序列化(其中最好的是Json.NET).要序列化对象:

var userData = new UserData {UserId = 0};
var userDataString = JsonConvert.SerializeObject(userData);

要反序列化对象:

var userData = JsonConvert.DeserializeObject<UserData>(userDataString);

序列化的JSON字符串如下所示:

{"UserId":0}

Answer 3

序列化和反序列化:

    public static T Deserialize<T>(this string toDeserialize)
    {
        XmlSerializer xmlSerializer = new XmlSerializer(typeof(T));
        using(StringReader textReader = new StringReader(toDeserialize))
        {      
            return (T)xmlSerializer.Deserialize(textReader);
        }
    }

    public static string Serialize<T>(this T toSerialize)
    {
        XmlSerializer xmlSerializer = new XmlSerializer(typeof(T));
        using(StringWriter textWriter = new StringWriter())
        {
            xmlSerializer.Serialize(textWriter, toSerialize);
            return textWriter.ToString();
        }
    }

Answer 4

代码安全说明

关于接受的答案,重要的是使用toSerialize.GetType()而不是typeof(T)在XmlSerializer构造函数中使用:如果使用第一个代码,则代码涵盖所有可能的方案,而使用后者则有时会失败.

这是一个链接,其中包含一些激励此语句的示例代码,XmlSerializer在typeof(T)使用时抛出一个Exception ,因为您将派生类型的实例传递给调用SerializeObject<T>()派生类型的基类中定义的方法:http:// ideone .com/1Z5J1.请注意,Ideone使用Mono来执行代码:使用Microsoft .NET运行时获得的实际异常消息与Ideone上显示的消息不同,但它失败的方式相同.

为了完整起见,我在这里发布完整的代码示例以供将来参考,以防Ideone(我发布代码的地方)将来变得不可用:

using System;
using System.Xml.Serialization;
using System.IO;

public class Test
{
    public static void Main()
    {
        Sub subInstance = new Sub();
        Console.WriteLine(subInstance.TestMethod());
    }

    public class Super
    {
        public string TestMethod() {
            return this.SerializeObject();
        }
    }

    public class Sub : Super
    {
    }
}

public static class TestExt {
    public static string SerializeObject<T>(this T toSerialize)
    {
        Console.WriteLine(typeof(T).Name);             // PRINTS: "Super", the base/superclass -- Expected output is "Sub" instead
        Console.WriteLine(toSerialize.GetType().Name); // PRINTS: "Sub", the derived/subclass

        XmlSerializer xmlSerializer = new XmlSerializer(typeof(T));
        StringWriter textWriter = new StringWriter();

        // And now...this will throw and Exception!
        // Changing new XmlSerializer(typeof(T)) to new XmlSerializer(subInstance.GetType()); 
        // solves the problem
        xmlSerializer.Serialize(textWriter, toSerialize);
        return textWriter.ToString();
    }
}

Answer 5

我的2p ......

        string Serialise<T>(T serialisableObject)
        {
            var xmlSerializer = new XmlSerializer(serialisableObject.GetType());

            using (var ms = new MemoryStream())
            {
                using (var xw = XmlWriter.Create(ms, 
                    new XmlWriterSettings()
                        {
                            Encoding = new UTF8Encoding(false),
                            Indent = true,
                            NewLineOnAttributes = true,
                        }))
                {
                    xmlSerializer.Serialize(xw,serialisableObject);
                    return Encoding.UTF8.GetString(ms.ToArray());
                }
            }
        }

Answer 6

public static string SerializeObject<T>(T objectToSerialize)
        {
            System.Runtime.Serialization.Formatters.Binary.BinaryFormatter bf = new System.Runtime.Serialization.Formatters.Binary.BinaryFormatter();
            MemoryStream memStr = new MemoryStream();

            try
            {
                bf.Serialize(memStr, objectToSerialize);
                memStr.Position = 0;

                return Convert.ToBase64String(memStr.ToArray());
            }
            finally
            {
                memStr.Close();
            }
        }

        public static T DerializeObject<T>(string objectToDerialize)
        {
            System.Runtime.Serialization.Formatters.Binary.BinaryFormatter bf = new System.Runtime.Serialization.Formatters.Binary.BinaryFormatter();
            byte[] byteArray = Convert.FromBase64String(objectToDerialize);
            MemoryStream memStr = new MemoryStream(byteArray);

            try
            {
                return (T)bf.Deserialize(memStr);
            }
            finally
            {
                memStr.Close();
            }
        }