title: Azure WebJob does not get response from PostAsync call
body: <p>I have taken a Windows Service, modified it to be a Console app, then Published it as a WebJob.  When it does a PostAsync command it gets no response AND does not execute the Console.WriteLine immediately following the PostAsync command.</p>
  
  <p>Here's the code:</p>
  
  <pre><code>            Console.WriteLine("Querying for access credentials...");
              HttpResponseMessage responsePut = null;
              try
              {
                  Console.WriteLine("PostAsync call");
                  responsePut = await client.PostAsync("/oauth/access_token", new StringContent("grant_type=client_credentials", Encoding.UTF8, "application/x-www-form-urlencoded"));
                  Console.WriteLine("Returned from PostAsync call");
              }
              catch (System.Net.Http.HttpRequestException err)
              {
                  Console.WriteLine("Unable to connect to the server (HttpRequestException)..." + err.InnerException);
                  throw new Exception("Unable to connect to the server (HttpRequestException)..." + err.InnerException);
              }
              catch (Exception err)
              {
                  Console.WriteLine("Unable to connect to the server (Exception)..." + err.Message + "::" + err.InnerException);
                  throw new Exception("Unable to connect to the server (Exception)..." + err.InnerException);
              }
  </code></pre>
  
  <p>The Azure Console Log is as follows:</p>
  
  <pre><code>[05/01/2018 05:45:31 &gt; e559bd: INFO] Querying for access credentials...
  [05/01/2018 05:45:31 &gt; e559bd: INFO] PostAsync call
  [05/01/2018 05:45:32 &gt; e559bd: SYS INFO] Status changed to Success
  [05/01/2018 05:45:32 &gt; e559bd: SYS INFO] Process went down, waiting for 60 seconds
  </code></pre>
  
  <p>This code works as a Windows Service and as a Windows Console App, but fails on Azure.  </p>
  
  <p>What can I do to determine what Azure is doing to me and/or suggestions to do this differently.</p>