title: Django REST Framework + Django REST Swagger + ImageField
body: <p>I created a simple Model with an ImageField and I wanna make an api view with django-rest-framework + django-rest-swagger, that is documented and is able to upload the file.</p>
  
  <p>Here is what I got:</p>
  
  <p><strong><code>models.py</code></strong></p>
  
  <pre><code>from django.utils import timezone
  from django.db import models
  
  class MyModel(models.Model):
  
      source = models.ImageField(upload_to=u'/photos')
      is_active = models.BooleanField(default=False)
      created_at = models.DateTimeField(default=timezone.now)
  
      def __unicode__(self):
          return u"photo {0}".format(self.source.url)
  </code></pre>
  
  <p><strong><code>serializer.py</code></strong></p>
  
  <pre><code>from .models import MyModel
  
  class MyModelSerializer(serializers.ModelSerializer):
  
      class Meta:
          model = MyModel
          fields = [
              'id',
              'source',
              'created_at',
          ]
  </code></pre>
  
  <p><strong><code>views.py</code></strong></p>
  
  <pre><code>from rest_framework import generics
  from .serializer import MyModelSerializer
  
  class MyModelView(generics.CreateAPIView):
      serializer_class = MyModelSerializer
      parser_classes = (FileUploadParser, )
  
      def post(self, *args, **kwargs):
          """
              Create a MyModel
              ---
              parameters:
                  - name: source
                    description: file
                    required: True
                    type: file
              responseMessages:
                  - code: 201
                    message: Created
          """
          return super(MyModelView, self).post(self, *args, **kwargs)
  </code></pre>
  
  <p><strong><code>urls.py</code></strong></p>
  
  <pre><code>from weddings.api.views import MyModelView
  
  urlpatterns = patterns(
      '',
      url(r'^/api/mymodel/$', MyModelView.as_view()),
  )
  </code></pre>
  
  <p>For me this should be pretty simple. However, I can't make the upload work. I always get this error response:
  <a href="https://i.stack.imgur.com/f0RFn.png" rel="noreferrer"><img src="https://i.stack.imgur.com/f0RFn.png" alt="enter image description here"></a></p>
  
  <p>I've read this part of the documentation from <a href="http://www.django-rest-framework.org/api-guide/parsers/#fileuploadparser" rel="noreferrer">django-rest-framework</a>:</p>
  
  <p><code>If the view used with FileUploadParser is called with a filename URL keyword argument, then that argument will be used as the filename. If it is called without a filename URL keyword argument, then the client must set the filename in the Content-Disposition HTTP header. For example Content-Disposition: attachment; filename=upload.jpg.</code></p>
  
  <p>However the Header is being passed by django-rest-swagger in the Request Payload property (from chrome console). </p>
  
  <p>If any more info is necessary, please let me know.</p>
  
  <p>I'm using <code>Django==1.8.8</code>, <code>djangorestframework==3.3.2</code> and <code>django-rest-swagger==0.3.4</code>.</p>