使用HttpWebRequest添加自定义标头

Question

我不确定这些突出显示的值是什么类型的标头，但我应该如何使用HttpWebRequest添加它们？

突出显示的部分是否被视为http请求或标题数据的正文？换句话说，哪种方式是正确的？

以下是我目前使用的代码：

HttpWebRequest request = (HttpWebRequest) WebRequest.Create("/securecontrol/reset/passwordreset");
request.Headers.Add("Authorization", "Basic asdadsasdas8586");
request.ContentType = "application/x-www-form-urlencoded";
request.Host = "www.xxxxxxxxxx.com";
request.Method = "POST";
request.Proxy = null;
request.Headers.Add("&command=requestnewpassword");
request.Headers.Add("&application=netconnect");

但是我应该使用以下代码来构建上面的Http请求吗？

string reqString = "&command=requestnewpassword&application=netconnect";
byte[] requestData = Encoding.UTF8.GetBytes(reqString);

HttpWebRequest request = (HttpWebRequest) WebRequest.Create("/securecontrol/reset/passwordreset");
request.Headers.Add("Authorization", "Basic ashAHasd87asdHasdas");
request.ContentType = "application/x-www-form-urlencoded";
request.ContentLength = requestData.Length;
request.Proxy = null;
request.Host = "www.xxxxxxxxxx.com";
request.Method = "POST";

using (Stream st = request.GetRequestStream())
st.Write(requestData, 0, requestData.Length);

Answer 1

创建服务，添加标头和阅读JSON响应的简单方法

private static void WebRequest()
{
    const string WEBSERVICE_URL = "<<Web Service URL>>";
    try
    {
        var webRequest = System.Net.WebRequest.Create(WEBSERVICE_URL);
        if (webRequest != null)
        {
            webRequest.Method = "GET";
            webRequest.Timeout = 20000;
            webRequest.ContentType = "application/json";
            webRequest.Headers.Add("Authorization", "Basic dcmGV25hZFzc3VudDM6cGzdCdvQ=");
            using (System.IO.Stream s = webRequest.GetResponse().GetResponseStream())
            {
                using (System.IO.StreamReader sr = new System.IO.StreamReader(s))
                {
                    var jsonResponse = sr.ReadToEnd();
                    Console.WriteLine(String.Format("Response: {0}", jsonResponse));
                }
            }
        }
    }
    catch (Exception ex)
    {
        Console.WriteLine(ex.ToString());
    }
}

Answer 2

恕我直言，它被视为格式错误的标题数据。

您确实希望将这些名称值对作为请求内容发送（这是POST的工作方式）而不是标题。

第二种方式是真的。

Answer 3

您应该执行 ex.StackTrace ，而不是 ex.ToString（）