用C＃进行json调用

Question

我正在尝试使用C＃进行json调用。我在创建呼叫时遇到了麻烦，但没有成功：

public bool SendAnSMSMessage(string message)
{
    HttpWebRequest request = (HttpWebRequest)
                             WebRequest.Create("http://api.pennysms.com/jsonrpc");
    request.Method = "POST";
    request.ContentType = "application/json";

    string json = "{ \"method\": \"send\", "+
                  "  \"params\": [ "+
                  "             \"IPutAGuidHere\", "+
                  "             \"[e-mail]\", "+
                  "             \"MyTenDigitNumberWasHere\", "+
                  "             \""+message+"\" " +
                  "             ] "+
                  "}";

    StreamWriter writer = new StreamWriter(request.GetRequestStream());
    writer.Write(json);
    writer.Close();

    return true;
}

任何有关如何使这项工作的建议，将不胜感激。

Answer 1

在您的代码中，您不会获得HttpResponse，因此您将看不到服务器端将您发回的内容。

您需要获得与获取（发出）请求类似的响应。所以

public static bool SendAnSMSMessage(string message)
{
  var httpWebRequest = (HttpWebRequest)WebRequest.Create("http://api.pennysms.com/jsonrpc");
  httpWebRequest.ContentType = "text/json";
  httpWebRequest.Method = "POST";

  using (var streamWriter = new StreamWriter(httpWebRequest.GetRequestStream()))
  {
    string json = "{ \"method\": \"send\", " +
                      "  \"params\": [ " +
                      "             \"IPutAGuidHere\", " +
                      "             \"[e-mail]\", " +
                      "             \"MyTenDigitNumberWasHere\", " +
                      "             \"" + message + "\" " +
                      "             ] " +
                      "}";

    streamWriter.Write(json);
  }
  var httpResponse = (HttpWebResponse)httpWebRequest.GetResponse();
  using (var streamReader = new StreamReader(httpResponse.GetResponseStream()))
  {
    var responseText = streamReader.ReadToEnd();
    //Now you have your response.
    //or false depending on information in the response
    return true;        
  }
}

我在pennysms文档中还注意到，他们期望内容类型为“ text / json”，而不是“ application /
json”。这可能没有什么区别，但是如果它不起作用，则值得尝试。