如何在 RedditSharp 中使用代理?

问题描述

我在脚本中使用来自 https://github.com/SirCmpwn/RedditSharp 的 RedditSharp我的，我只是问，使用它连接时如何实现代理?我可以更改代理中标吗?

I'm using RedditSharp from https://github.com/SirCmpwn/RedditSharp in a script of mine, and I'm simply asking, when connecting using this how do I implement a proxy? and could I change the proxy midscript?

推荐答案

没有独立的方法，不修改这个库源代码是无法完成的.

There no standalone way, you can't accomplish this without modifying this library source code.

最无痛的方式:

1. RedditSharp 的重载构造函数 - 添加以 IWebAgent 作为类型的新参数.所以它看起来像这样:

1. Overload constructor of RedditSharp - add new argument with IWebAgent as type. So it will look like this:

public Reddit() : this(new WebAgent())
{

}

public Reddit(IWebAgent agent)
{
    JsonSerializerSettings = new JsonSerializerSettings();
    JsonSerializerSettings.CheckAdditionalContent = false;
    JsonSerializerSettings.DefaultValueHandling = DefaultValueHandling.Ignore;
    _webAgent = agent;
    CaptchaSolver = new ConsoleCaptchaSolver();
}

从 RedditSharp.WebAgent 类声明中删除密封"关键字.

Remove "sealed" keyword from RedditSharp.WebAgent class declaration.

使 RedditSharp.WebAgent.CreateRequest 方法成为虚拟方法，所以它看起来像这样:

Make RedditSharp.WebAgent.CreateRequest method virtual, so it will look like this:

public virtual HttpWebRequest CreateRequest(string url, string method, bool prependDomain = true)
{
    ...
}

基于旧版创建您自己的 WebAgent:

Create your own WebAgent based on old-one:

public class MyAgent: WebAgent
{
    public IWebProxy Proxy { get; set; }

    public override HttpWebRequest CreateRequest(string url, string method, bool prependDomain = true)
    {
        var base_request = base.CreateRequest(url, method, prependDomain);

        if (Proxy != null)
        {
            base_request.Proxy=Proxy;
        }

        return base_request;
    }
}

在您的代码中使用它:

Use it in your code:

var agent = new MyAgent();
var reddit = new Reddit(agent);

...

agent.Proxy = new WebProxy("someproxy.net", 8080);

所以现在您可以随时随地设置代理.没有真正测试过，但它必须工作.

So now you can set proxy anytime, from anywhere. Not tested really, but it must work.

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