title: Getting the Django-generated unique name of a file uploaded multiple times
body: <p>I am using DRF backend to upload files. In my specific case I will want to get the name of the file, <strong>after</strong> it has been uploaded. The reason is that if a user uploads a file with same name, I am still able to process it independently.</p>
  
  <p><strong>views.py</strong>:</p>
  
  <pre><code>class ImageUploaderView(viewsets.ModelViewSet):
      renderer_classes = [renderers.JSONRenderer]
      queryset = ImageUploader.objects.all()
      serializer_class = ImageUploaderSerializer
      parser_classes = (MultiPartParser,)
  </code></pre>
  
  <p><strong>serializer.py</strong>:</p>
  
  <pre><code>class ImageUploaderSerializer(serializers.ModelSerializer):
      class Meta:
          model = ImageUploader
  </code></pre>
  
  <p><strong>models.py</strong>:</p>
  
  <pre><code>class ImageUploader(models.Model):
      # name=models.ImageField(upload_to='media')
      name=models.FileField(upload_to='media')
  </code></pre>
  
  <p>I tried to put signals and hooks after the model definitions but I am not being able to get this filename. Can someone shed some light pls?</p>
  
  <p><strong>UPDATE:</strong> Let me elaborate what I want to achieve essentially:</p>
  
  <ul>
  <li><p>User1 hits endpoint "/api/calculate_interest_rate" which is rendered
  by a frontend React component. "calculate_interest_rate" is served by
  DRF, and lets the user upload a CSV file. This will be stored as
  "user1.csv", the file is processed and then tabulated (rendered by
  React).</p></li>
  <li><p>At the same time and in parallel to User1, User2 hits the same endpoint "/api/calculate_interest_rate" and
  by mistake he saves his file as "user1.csv", and uploads it to the systemn.</p></li>
  </ul>
  
  <p>So I want to be able to detect both names of the file in order to process it. By using always the same default filename (ex. using the OverwriteStorage() technique), I will probably cause chaos when two or more users are using the same filename. Therefore I am looking into a technique that allows me to get the filename as is, and process it immediately.</p>