如何快速保存/加载类的实例到文件 [英] How to quickly save/load class instance to file

问题描述

我在我的应用程序的类/结构的几个系列。

I have several collections of classes/structs in my app.

类只是一个类的字段

class A
{
  public int somevalue;
  public string someothervalue
}

和我的收藏

List<A> _myList;

我需要能够保存_myList和负载。我只是想保存所有类字段到文件并加载。我不想花时间写我自己的保存/载入。是否有.NET任何工具来帮助我。我不关心文件格式。

I need to be able to save _myList and load. I just want to save all class fields to file and load. I don't want to spend time writing my own save/load. Are there any tools in .NET to help me. I don't care about the file format.

推荐答案

XMLSerializer的是不是很难使用。只要你的对象并不大，这是pretty的快。序列化出一些巨大的物体在我的几个应用程序。它永远和生成的文件几乎是100兆，但他们可编辑的我应该需要调整一些东西。再加上它并不重要，如果我将字段添加到我的对象。对象的旧版本的序列化的文件仍然反序列化正确。我做一个单独的线程序列化，因此它不会不管需要我的情况有多长。需要说明的是，你的 A 类必须有一个构造XMLSerialziation工作。

XMLSerializer isn't hard to use. As long as your objects aren't huge, it's pretty quick. I serialize out some huge objects in a few of my apps. It takes forever and the resulting files are almost 100 megs, but they're editable should I need to tweak some stuff. Plus it doesn't matter if I add fields to my objects. The serialized files of the older version of the object still deserialize properly.. I do the serialization on a separate thread so it doesn't matter how long it takes in my case. The caveat is that your A class has to have a constructor for XMLSerialziation to work.

下面是一些工作code我用序列化/反序列化的错误处理拆出来的可读性...

Here's some working code I use to serialize/deserialize with the error handling ripped out for readibility...

private List<A> Load()
{
    string file = "filepath";
    List<A> listofa = new List<A>();
    XmlSerializer formatter = new XmlSerializer(A.GetType());
    FileStream aFile = new FileStream(file, FileMode.Open);
    byte[] buffer = new byte[aFile.Length];
    aFile.Read(buffer, 0, (int)aFile.Length);
    MemoryStream stream = new MemoryStream(buffer);
    return (List<A>)formatter.Deserialize(stream);
}


private void Save(List<A> listofa)
{
    string path = "filepath";
    FileStream outFile = File.Create(path);
    XmlSerializer formatter = new XmlSerializer(A.GetType());
    formatter.Serialize(outFile, listofa);
}

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