title: django celery get() for getting the result form task
body: <p>I have an API call to the server where I am doing some processing using celery. The output from the celery is passed back to client.</p>
  
  <p>I used </p>
  
  <pre><code>som = task_async_get_cached_session.delay('session_123')
  print 'before'
  result = som.get(timeout=1)
  print 'after'
  </code></pre>
  
  <p>I looked at the <code>celery -l INFO</code> and the task get succeeded in 
  <code>succeeded in 0.024465521s: {data: {session_123: something}}</code></p>
  
  <p>But my client was not get response back and and I put timeout and now I am getting <code>The operation timed out. (&lt;class 'celery.exceptions.TimeoutError'&gt;)</code></p>
  
  <p>When I looked at the logs trace, only <code>before</code> is printed. My question how to use celery without <code>get()</code> when you want the result generated from the celery task to the client?</p>
  
  <p>I know that <code>get()</code> will convert my API call to synchronous, but what alternate I have ? What is the real use of celery ? Is there any other way to change my API to behave Asynchronous?  </p>