Django Rest Framework: Assign image to model field from uploaded folder - django-models

Question

I am HTTP posting a .zip File to my Django app via Django - Rest - Framework. This zip File is a Folder that contains several files, among them an image. I would like to extract the folder once its uploaded, select the image and assign it to the model. Is this possible? If not, maybe I can write a property that gets the thumbnail image? I want to be able to show all the thumbnails in a gallery later.
Something like:
class FUploadSerializer(serializers.ModelSerializer):
file = serializers.FileField()
class Meta:
model = FUpload
fields = ('created','file')
class FUpload(models.Model):
created = models.DateTimeField(auto_now_add=True)
file = models.FileField(upload_to=get_path)
thumbnail = ImageField() ??? # get from uploaded folder
EDIT
I tried the following, but am getting: "image": [
"The submitted data was not a file. Check the encoding type on the form."
] . Am I thinking about this completely backwards? Is it "better" to let the client handle sending the image and .zip folder separately (data is sent from a script)? Is the ModelViewSet the right place to handle this?
class Thumbnail(models.Model):
image = models.ImageField()
file = models.FileField()
class ThumbnailViewSet(viewsets.ModelViewSet):
queryset = Thumbnail.objects.all()
serializer_class = ThumbnailSerializer
parser_classes = (MultiPartParser, FormParser,)
def perform_create(self, serializer):
file = self.request.FILES['file']
zf = zipfile.ZipFile(file)
content_list = zf.namelist()
imgdata = zf.open(content_list[0])
serializer.save(image=imgdata)
class ThumbnailSerializer(serializers.ModelSerializer):
file = serializers.FileField()
image = serializers.ImageField(allow_empty_file=True, required=False)
class Meta:
model = Thumbnail
fields = ('file', 'image')

Answer 1

In order to make this work, you are going to need to implement your own custom parser that can handle the incoming zip file, extract it, and assign some of the files to specific fields.
class ZipParser(FileUploadParser):
"""
Parses an incoming ZIP file into separate fields.
The ZIP file is expected to be the only data passed in the request,
use the `MultiPartParser` as a base if it will be passed in using form
data instead.
"""
media_type = 'application/zip' # only needed for `FileUploadParser`
def parse(self, stream, media_type=None, parser_context=None):
import io
import zipfile
parsed = super(ZipParser, self).parse(
self,
media_type=media_type,
parser_context=parser_context
)
data = parsed.files
data['zip_file'] = data['file'] # Keep the original zip file as `zip_file`
parsed_file = data['file'] # Also store a reference to the original file so it can be used
# Try to parse the zip file
# This should also handle proper file handling (such as not passing a zip file)
with zipfile.ZipFile(parsed_file) as zip_file:
# Get the `bytes` object for a file using `read(file_name)`
image_data = zip_file.read('image.png') # Replace `image.png` with the image file
file_data = zip_file.read('info.txt')
# DRF expects that files are using an IO compatible object, so we need `BytesIO` here
image_stream = io.BytesIO(image_data)
file_stream = io.BytesIO(file_data)
# The changed files should go back onto the files dictionary
data['image'] = image_stream
data['file'] = file_stream
return parsed
You can alternatively do this outside of a parser before the data is passed into a serializer if you are using your own custom view.