我一直在使用的一个应用程序在尝试序列化类型时失败了。

像这样的陈述

XmlSerializer lizer = new XmlSerializer(typeof(MyType));

生产:

System.IO.FileNotFoundException occurred
  Message="Could not load file or assembly '[Containing Assembly of MyType].XmlSerializers, Version=1.0.0.0, Culture=neutral, PublicKeyToken=null' or one of its dependencies. The system cannot find the file specified."
  Source="mscorlib"
  FileName="[Containing Assembly of MyType].XmlSerializers, Version=1.0.0.0, Culture=neutral, PublicKeyToken=null"
  FusionLog=""
  StackTrace:
       at System.Reflection.Assembly._nLoad(AssemblyName fileName, String codeBase, Evidence assemblySecurity, Assembly locationHint, StackCrawlMark& stackMark, Boolean throwOnFileNotFound, Boolean forIntrospection)
       at System.Reflection.Assembly.nLoad(AssemblyName fileName, String codeBase, Evidence assemblySecurity, Assembly locationHint, StackCrawlMark& stackMark, Boolean throwOnFileNotFound, Boolean forIntrospection)

我没有为我的类定义任何特殊的序列化器。

我该如何解决这个问题?


当前回答

看到过很多使用ConcurrentDictionary的建议,但没有可靠的例子,所以我将全身心投入到这个解决方案竞赛中。我不是一个线程安全的开发人员,所以如果这段代码不可靠,请为那些跟随的人说出来。

public static class XmlSerializerHelper
{
    private static readonly ConcurrentDictionary<Type, XmlSerializer> TypeSerializers = new ConcurrentDictionary<Type, XmlSerializer>();

    public static XmlSerializer GetSerializer(Type type)
    {
        return TypeSerializers.GetOrAdd(type,
        t =>
        {
            var importer = new XmlReflectionImporter();
            var mapping = importer.ImportTypeMapping(t, null, null);
            return new XmlSerializer(mapping);
        });
    }
}

我看过其他关于ConcurrentDictionary和Lazy加载值的文章。我不确定这是否与此相关,但这里是它的代码:

private static readonly ConcurrentDictionary<Type, Lazy<XmlSerializer>> TypeSerializers = new ConcurrentDictionary<Type, Lazy<XmlSerializer>>();

public static XmlSerializer GetSerializer(Type type)
{
    return TypeSerializers.GetOrAdd(type,
    t =>
    {
        var importer = new XmlReflectionImporter();
        var mapping = importer.ImportTypeMapping(t, null, null);
        var lazyResult = new Lazy<XmlSerializer>(() => new XmlSerializer(mapping), LazyThreadSafetyMode.ExecutionAndPublication);
        return lazyResult;
    }).Value;
}

其他回答

您的类型可能会引用在GAC和本地bin文件夹中都找不到的其他程序集==>…

或其依赖项之一。该系统 无法找到指定的文件"

您能举例说明要序列化的类型吗?

注意:确保你的类型实现Serializable。

在我的一个。net标准dll中也有类似的问题。

我使用了Microsoft.XmlSerializer.Generator nuget,它可以在. net Core和. net Standard上预生成XmlSerializer。

Like Martin Sherburn said, this is normal behavior. The constructor of the XmlSerializer first tries to find an assembly named [YourAssembly].XmlSerializers.dll which should contain the generated class for serialization of your type. Since such a DLL has not been generated yet (they are not by default), a FileNotFoundException is thrown. When that happenes, XmlSerializer's constructor catches that exception, and the DLL is generated automatically at runtime by the XmlSerializer's constructor (this is done by generating C# source files in the %temp% directory of your computer, then compiling them using the C# compiler). Additional constructions of an XmlSerializer for the same type will just use the already generated DLL.

更新:从。net 4.5开始,XmlSerializer不再执行代码生成,也不再为了在运行时创建序列化程序集而使用c#编译器执行编译,除非通过设置配置文件设置(useLegacySerializerGeneration)显式强制执行。此更改消除了对csc.exe的依赖,并提高了启动性能。来源:. net Framework 4.5 Readme,章节1.3.8.1。

异常由XmlSerializer的构造函数处理。你不需要自己做任何事情,你可以点击“继续”(F5)继续执行你的程序,一切都会好的。如果你被异常停止程序执行并弹出异常助手所困扰,你要么关闭“只是我的代码”,要么将FileNotFoundException设置为在抛出时中断执行,而不是在“User-unhandled”时。

打开“Just My Code”,请转到“Tools >> Options >> Debugging >> General >> enable Just My Code”。要关闭FileNotFound被抛出时的执行中断,请转到调试>>异常>>找到>>输入'FileNotFoundException' >>取消System.IO.FileNotFoundException中的' thrown '复选框。

此异常也可以由名为BindingFailure的托管调试助手(MDA)捕获。

如果您的应用程序被设计为附带预构建序列化程序集,则此MDA非常有用。这样做是为了提高应用程序的性能。它允许我们确保预先构建的序列化程序集由构建过程正确构建,并由应用程序加载,而无需动态重新构建。

除了在这种情况下,它实际上没什么用,因为正如其他帖子所说,当绑定错误被Serializer构造函数捕获时,序列化程序集将在运行时重新构建。所以你可以把它关掉。

要序列化的自定义类:

[Serializable]
public class TestClass
{
    int x = 2;
    int y = 4;
    public TestClass(){}
    public TestClass(int x, int y)
    {
        this.x = x;
        this.y = y;
    }

    public int TestFunction()
    {
        return x + y;
    }
}

我已经附上了代码片段。也许这个能帮到你。

static void Main(string[] args)
{
    XmlSerializer xmlSerializer = new XmlSerializer(typeof(TestClass));

    MemoryStream memoryStream = new MemoryStream();
    XmlTextWriter xmlWriter = new XmlTextWriter(memoryStream, Encoding.UTF8);

    TestClass domain = new TestClass(10, 3);
    xmlSerializer.Serialize(xmlWriter, domain);
    memoryStream = (MemoryStream)xmlWriter.BaseStream;
    string xmlSerializedString = ConvertByteArray2Str(memoryStream.ToArray());

    TestClass xmlDomain = (TestClass)DeserializeObject(xmlSerializedString);

    Console.WriteLine(xmlDomain.TestFunction().ToString());
    Console.ReadLine();
}