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.
Here is what I got:
models.py
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) serializer.py
from .models import MyModel class MyModelSerializer(serializers.ModelSerializer): class Meta: model = MyModel fields = [ 'id', 'source', 'created_at', ] views.py
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) urls.py
from weddings.api.views import MyModelView urlpatterns = patterns( '', url(r'^/api/mymodel/$', MyModelView.as_view()), ) For me this should be pretty simple. However, I can't make the upload work. I always get this error response:
I've read this part of the documentation from django-rest-framework:
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.
However the Header is being passed by django-rest-swagger in the Request Payload property (from chrome console).
If any more info is necessary, please let me know.
I'm using Django==1.8.8, djangorestframework==3.3.2 and django-rest-swagger==0.3.4.
2 Answers
Answers 1
I got this working by making a couple of changes to your code.
First, in models.py, change ImageField name to file and use relative path to upload folder. When you upload file as binary stream, it's available in request.data dictionary under file key (request.data.get('file')), so the cleanest option is to map it to the model field with the same name.
from django.utils import timezone from django.db import models class MyModel(models.Model): file = 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.file.url) In serializer.py, rename source field to file:
class MyModelSerializer(serializers.ModelSerializer): class Meta: model = MyModel fields = ('id', 'file', 'created_at') In views.py, don't call super, but call create():
from rest_framework import generics from rest_framework.parsers import FileUploadParser from .serializer import MyModelSerializer class MyModelView(generics.CreateAPIView): serializer_class = MyModelSerializer parser_classes = (FileUploadParser,) def post(self, request, *args, **kwargs): """ Create a MyModel --- parameters: - name: file description: file required: True type: file responseMessages: - code: 201 message: Created """ return self.create(request, *args, **kwargs) I've used Postman Chrome extension to test this. I've uploaded images as binaries and I've manually set two headers:
Content-Disposition: attachment; filename=upload.jpg Content-Type: */* Answers 2
It has been my experience that the FileUploadParser works with this format of a request:
curl -X POST -H "Content-Type:multipart/form-data" \ -F "file=@{filename};type=image/jpg" \ https://endpoint.com/upload-uri/ The request.data['file'] in your view will have the file.
Maybe if you try a Content-Type:multipart/form-data header, you will have luck.
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