Django REST Framework image upload

Question

I have model Product:

def productFile(instance, filename):
    return '/'.join( ['products', str(instance.id), filename] )

class Product(models.Model):
    ...

    image = models.ImageField(
        upload_to=productFile,
        max_length=254, blank=True, null=True
    )
    ...

Then I have serializer:

class ProductSerializer(serializers.ModelSerializer):
    class Meta:
        model = Product
        fields = (
            ...
            'image',
            ...
        )

And then I have views:

class ProductViewSet(BaseViewSet, viewsets.ModelViewSet):
    queryset = Product.objects.all()
    serializer_class = ProductSerializer

How I can upload image with Postman? What is the best practices to upload image to model? Thank you.

Remove the paranthesis, serializer_class = ProductSerializer.. If you are using ModelViewSet, then you don't need to write anything more... ViewSet would handle pretty much everything for you. — zaidfazil
– zaidfazil, Commented Aug 8, 2017 at 9:51
Ok, thanks, and how can I upload image with postman, or some Ajax request? — Danila Kulakov
– Danila Kulakov, Commented Aug 8, 2017 at 10:47
When I use that in my model, I get a ValueError("Cannot serialize function: lambda") — Rony Azrak
– Rony Azrak, Commented Dec 18, 2018 at 6:00
@Rony Azrak, I'he just updated the question with a solution that works. use method productFile — Danila Kulakov
– Danila Kulakov, Commented Dec 18, 2018 at 16:12
Can you send me your code to post data in model with image file I'm wondering for it..my code show me bad request 400 http://.... — Muhammad Abdullah
– Muhammad Abdullah, Commented Aug 22, 2020 at 7:14

Shehab ElDin · Accepted Answer · 2017-08-08 11:32:10Z

34

you can create separate endpoint for uploading images, it would be like that:

class ProductViewSet(BaseViewSet, viewsets.ModelViewSet):
    queryset = Product.objects.all()
    serializer_class = ProductSerializer

    @detail_route(methods=['post'])
    def upload_docs(request):
        try:
            file = request.data['file']
        except KeyError:
            raise ParseError('Request has no resource file attached')
        product = Product.objects.create(image=file, ....)

you can go around that solution

-- update: this's how to upload from postman

edited Aug 8, 2017 at 11:32

answered Aug 8, 2017 at 11:09

Shehab ElDin

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6 Comments

Shehab ElDin Over a year ago

you can upload using postman by select form-data on body and select type to file and click on choose file, write file on key field see the update above.

Danila Kulakov Over a year ago

Thank you! Can I have url with method "PUT" like /api/products/33/upload_docs ?

Shehab ElDin Over a year ago

yes you can of course but I didn't check if you can upload with PUT method or not

Danila Kulakov Over a year ago

hm, I'm, just tried your solution, but I got error {"detail":"Missing filename. Request should include a Content-Disposition header with a filename parameter."} . Do you know how can I "turn off" this requirement? Once more Thank you! :)

Danila Kulakov Over a year ago

problem was with parser, I used FileUploadParser instead of MultiPartParser :) Thank you!

|

Radesh · Accepted Answer · 2023-07-16 08:01:30Z

24

I lately start Django and have same problem for upload image.

All steps that i done

Install Pillow for using ImageField
```
pip install Pillow
```

In Settings.py add these lines

MEDIA_ROOT = os.path.join(BASE_DIR, 'media')
MEDIA_URL = '/media/' # 'http://myhost:port/media/'

Use ImageField in model.py (create nameFile function for create folder and name of file)

def upload_to(instance, filename):
    return '/'.join(['images', str(instance.name), filename])

class UploadImageTest(models.Model):
    name = models.CharField(max_length=100)
    image = models.ImageField(upload_to=upload_to, blank=True, null=True)

serializer.py

class ImageSerializer(serializers.ModelSerializer):
    class Meta:
        model = UploadImageTest
        fields = ('name', 'image')

views.py

class ImageViewSet(ListAPIView):
    queryset = UploadImageTest.objects.all()
    serializer_class = ImageSerializer

    def post(self, request, *args, **kwargs):
        file = request.data['file']
        image = UploadImageTest.objects.create(image=file)
        return HttpResponse(json.dumps({'message': "Uploaded"}), status=200)

urls.py: add this line

path('upload/', views.ImageViewSet.as_view(), name='upload'),

admin.py: add this line (for check in admin)
```
admin.site.register(UploadImageTest)
```

in terminal

python manage.py makemigrations
python manage.py migrate

edited Jul 16, 2023 at 8:01

answered Mar 17, 2019 at 9:35

Radesh

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2 Comments

Danila Kulakov Over a year ago

Have you tried to use MultiPartParser in views? Also I would recommend you to make new endpoint to 'upload_image' in ImageViewSet

OurangZeb Khan Over a year ago

can you tell how to update this in request.

stephendwolff · Accepted Answer · 2022-05-30 01:11:47Z

models.py

from django.db import models

class ImageUpload(models.Model):
    title = models.CharField(max_length=50)
    images = models.ImageField('images')

serializers.py

from rest_framework import serializers
from .models import ImageUpload

class ImageUploadSerializer(serializers.HyperlinkedModelSerializer):

    class Meta:
        model = ImageUpload
        fields= (
            'title',
            'images'
        )

views.py

from rest_framework import viewsets
from .models import ImageUpload
from .serializers import ImageUploadSerializer

class ImageUploadViewSet(viewsets.ModelViewSet):
    queryset = ImageUpload.objects.all()
    serializer_class = ImageUploadSerializer

urls.py

from django.urls import path, include
from rest_framework.routers import DefaultRouter
from . import views

router = DefaultRouter()
router.register(r'imageupload', views.ImageUploadViewSet)

urlpatterns = [
    path('imageupload', include(router.urls)),
]

Adi · Accepted Answer · 2022-03-30 03:49:02Z

1

this worked for me

class UserAvatarUpload(ListAPIView):
    
    parser_classes = [MultiPartParser, FormParser]
    serializer_class = ImageSerializer

    def post(self, request, *args, **kwargs):
        username = request.data['username']
        file = request.data['image']
        user = MyModel.objects.get(username=username)
        user.image = file
        user.save()
        return Response("Image updated!", status=status.HTTP_200_OK)

answered Mar 30, 2022 at 3:49

Adi

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