Question

我想将多个帖子请求发送到网站。我搜索了Google并获得了此代码：

 class Program
{
    public static HttpWebRequest HttpWReq = (HttpWebRequest)WebRequest.Create("http://example.com/login.php");
    public static ASCIIEncoding encoding = new ASCIIEncoding();
    public static string postData = "";
    public static bool first = true;
    static void Main(string[] args)
    {
        string uname = "", pass = "", Final = "";

        while (1>0)
        {
            Console.WriteLine("Enter uname ,then password");
            uname = Console.ReadLine();
            pass = Console.ReadLine();
            if (uname == "0")
                break;
            Final = letstry(uname, pass);

            Console.WriteLine(Final);
            Console.WriteLine("Finish That");
        }
    }

    public static string letstry(string uname, string pass)
    {  
        postData = "uname=" + uname;
        postData += ("&pass=" + pass);
        byte[] data = encoding.GetBytes(postData);
        if (first)
        {
            HttpWReq.Method = "POST";
            HttpWReq.ContentType = "application/x-www-form-urlencoded";
            HttpWReq.ContentLength = data.Length;
            first = !first;
        }

        Stream newStream = HttpWReq.GetRequestStream();
        newStream.Write(data, 0, data.Length);
        newStream.Close();

        WebResponse resp = HttpWReq.GetResponse();
        StreamReader sr = new System.IO.StreamReader(resp.GetResponseStream());
        return sr.ReadToEnd().Trim();
    }
}

我收到连接已关闭的错误（newstream）。为什么我不能使用相同的连接发送多个请求？

我能想到的唯一想法是将流变量发送到letstry，而不是创建newstream。

我不是专家，所以我很抱歉任何不必要的错误。

tyvm求助：）

Answer 1

WebRequest的目的是做它所说的：提出一个请求。如果您想要发出多个请求，请每次都创建新的WebRequest。

如果KeepAlive属性为true，则请求尝试使用相同的连接（如果可能）。有关详细信息，请参阅Understanding System.Net Connection Management and ServicepointManager。

发送多个帖子请求

1 个答案: