WebException重新抛出

Question

下面是代码：

    public static string Post(string requestUriString, string s)
    {
        var request = (HttpWebRequest)WebRequest.Create(requestUriString);
        var data = Encoding.ASCII.GetBytes(s);
        request.Method = "POST";
        request.ContentType = "application/x-www-form-urlencoded";
        request.ContentLength = data.Length;

        using (var stream = request.GetRequestStream())
        {
            stream.Write(data, 0, data.Length);
        }

        try
        {
            var response = (HttpWebResponse)request.GetResponse();
            return new StreamReader(response.GetResponseStream()).ReadToEnd();
        }
        catch (WebException webException)
        {
            Console.WriteLine(webException);
            throw;
        }
    }

当程序抛出异常时，try / catch块将正常工作。

然而，当我尝试重新抛出异常时，程序崩溃了。

Answer 1

它应该会崩溃，因为你没有处理异常重新抛出，你需要在Main()方法中try catch，如下所示。

static void Main(string[] args)
{
    try
    {
        Post("https://stackoverflow.com/", "test");
    }
    catch (Exception)
    {
        //  Handle exception re-throw,
    }
}

Answer 2

这样做：

class Program {
    static void Main(string[] args) {
        var response = Post(...);

        if(response.Status == "success") {
            //ok
        } else {
            //request failed
        }
    }

    public static string Post(string requestUriString, string s) {
        var request = (HttpWebRequest)WebRequest.Create(requestUriString);
        var data = Encoding.ASCII.GetBytes(s);
        request.Method = "POST";
        request.ContentType = "application/x-www-form-urlencoded";
        request.ContentLength = data.Length;

        using (var stream = request.GetRequestStream()) {
            stream.Write(data, 0, data.Length);
        }

        try {
            var response = (HttpWebResponse)request.GetResponse();
            var body = new StreamReader(response.GetResponseStream()).ReadToEnd();
            return new Response("success", body);
        } catch (WebException webException) {
            Console.WriteLine(webException);
            return new Response("failed", "");
        }
    }

    class Response {
        public string Status { get; private set; }
        public string Body { get; private set; }

        public Response(string status, string response) {
            Status = status;
            Body = response;
        }
    }
}

根据您的需求和应用程序大小，您可以基于http响应码(200/404/500...)将状态设置为枚举。等，并添加一些更多的处理，因为我发布的示例不应该用于更大或生产关键的东西。

你提到这是一个控制台应用程序，如果它只做一个单一的请求，而不做任何其他事情，那么它应该可以很好地完成这项工作。

Answer 3

使用此代码解决的问题：

using System.Net;
using System.Text;
using System.IO;
using System;

namespace Botnet
{
    class Program
    {
        static void Main(string[] args)
        {
            try
            {
                Console.WriteLine(Post("http://www.example.com/recepticle.aspx", "thing1=hello&thing2=world"));
            }
            catch (WebException webException)
            {
                Console.WriteLine(webException);
            }
            finally
            {
                Console.ReadLine();
            }
        }

        public static string Post(string requestUriString, string s)
        {
            var request = (HttpWebRequest)WebRequest.Create(requestUriString);
            var data = Encoding.ASCII.GetBytes(s);
            request.Method = "POST";
            request.ContentType = "application/x-www-form-urlencoded";
            request.ContentLength = data.Length;

            using (var stream = request.GetRequestStream())
            {
                stream.Write(data, 0, data.Length);
            }

            try
            {
                var response = (HttpWebResponse)request.GetResponse();
                return new StreamReader(response.GetResponseStream()).ReadToEnd();
            }
            catch (WebException webException)
            {
                Console.WriteLine(webException);
                throw;
            }
        }
    }
}