Django REST Framework + Django REST Swagger + ImageField - python

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.

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: */*

This is the final solution I came up with:
from rest_framework import generics
from rest_framework.parsers import FormParser, MultiPartParser
from .serializer import MyModelSerializer
class MyModelView(generics.CreateAPIView):
serializer_class = MyModelSerializer
parser_classes = (FormParser, MultiPartParser)
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)
All I had to do was change the parsers from FileUploadParser to (FormParser, MultiPartParser)

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.

Related

django read mp3 file and send it as response

views.py
from . import models, serializers
from rest_framework import viewsets, status
from rest_framework.response import Response
from rest_framework.views import APIView
class getSongData(APIView):
serializer_class=serializers.SongSerializer
def get(self, request, id, format=None):
serializer = serializers.SongSerializer(models.Song.objects.get(id=id))
file_loc = serializer.data['audio_file'] # go below to see the data
# read the mp3 file
return Response(file_data)
urls.py
from django.urls import path
from . import views
urlpatterns = [
path('songs/audio/<int:id>', views.getSongData.as_view(), name='audio')
]
serializers.py
from rest_framework import serializers
from . import models
class SongSerializer(serializers.ModelSerializer):
class Meta:
model = models.Song
fields = '__all__'
models.py
from django.db import models
from datetime import datetime
class Song(models.Model):
title = models.CharField(max_length=64)
audio_file = models.FileField()
genre = models.CharField(max_length=64)
created_at = models.DateTimeField(default=datetime.utcnow)
The data
[
{
"id": 1,
"title": "Kubbi | Cascade",
"audio_file": "/media/Kubbi__Cascade.mp3",
"genre": "Instrumental",
"created_at": "2021-07-24T10:21:48Z"
}
]
When the user clicks on a song (lets say the song's id=1), a request gets sent to 'http://localhost:8000/api/songs/audio/1' then in views.py I extract the song's location via serializer.data['audio_file'] which is = "/media/Kubbi__Cascade.mp3", all i want to do is to read this audio file and send the data as a Response back to the frontend, I tried many solutions but they were throwing errors...
if you want to do it using pure django this would work
from django.http import FileResponse
class getSongData(APIView):
serializer_class=serializers.SongSerializer
def get(self, request, id, *args, **kwargs):
song = models.Song.objects.get(id=id)
return FileResponse(song.audio_file.open())
but preferably you should serve files through a reverse proxy for better performance.

Creat REST Api Using Django Rest Framework

I am trying to create a Restful Api for following methods to run jenkins jobs to run on saucelabs. I wanna queue jobs using restful API. I am using Django Restful Framework.
CreateMethod :
Accepts two fileds: ProjectName and URL
and returns a Token ID.
VerifyStatus:
Accepts Token ID and returns three fields. TokenID, running:True/False and
no_of_jobs: integervalue (0 if Not specified)
relseaseMethod:
Accepts release token call and returns true if success.
I am new to Restful API with, I am trying to run Jenkins job on sauce lab and queue them using a restful api on python Djano restframework. Guide me through.
Views.py
class RegisterTestMethodView(APIView):
authentication_classes = [SessionAuthentication, TokenAuthentication, BasicAuthentication]
permission_classes = [IsAuthenticated] #No access (not even read if not authorized)
def post(self, request, format=None):
serializer = RegisterTestMethodSerializers(data=request.data)
if serializer.is_valid():
serializer.save()
return Response({'tokenid':serializer.data['id']}, status=status.HTTP_201_CREATED)
return Response(serializer.errors, status=status.HTTP_400_BAD_REQUEST)
class CheckStatusView(APIView):
def get_object(self, pk):
try:
return Jobs.objects.get(pk=pk)
except Jobs.DoesNotExist:
raise Http404
def get(self, request, pk, format=None):
snippet = self.get_object(pk)
serializer = RegisterTestMethodSerializers(snippet)
return Response({"tokenid":serializer.data["id"], "Runtestnow" : False, "VMcount" : 0 })
class ReleaseTokenView(APIView):
def get_object(self, pk):
try:
return Jobs.objects.get(pk=pk)
except Jobs.DoesNotExist:
raise Http404
def delete(self, request, pk, format=None):
snippet = self.get_object(pk)
snippet.delete()
return Response(data={'Deleted':True}, status=status.HTTP_204_NO_CONTENT)
Serailizers.py
rom rest_framework import serializers
from .models import Jobs
from random import random
RegisterTestMethodSerializers(serializers.HyperlinkedModelSerializer):
class Meta:
model = Jobs
fields = ('id','name','url')
Models.py
from django.db import models
# Create your models here.
class Jobs(models.Model):
name = models.CharField(max_length=100)
url = models.URLField()
def __str__(self):
return self.name
Urls.py
from django.urls import path, include
from . import views
from .views import (RegisterTestMethodView,
RegisterTestMethodViewDetail,
CheckStatusView,
ReleaseTokenView
)
from rest_framework import routers
from rest_framework.authtoken.views import obtain_auth_token
from rest_framework_simplejwt.views import TokenObtainPairView, TokenRefreshView
urlpatterns = [
path('', include(router.urls)),
path('registertestmethod/',RegisterTestMethodView.as_view()),
path('registertestmethod/<int:pk>/',
RegisterTestMethodViewDetail.as_view()),
path('checkstatus/<int:pk>', CheckStatusView.as_view()),
path('releasetoken/<int:pk>', ReleaseTokenView.as_view()),
]
I have addded my Code here. I have other classes and function in my project as well.I tried to delete all of those. You might see extra import as a result of that. My code does following things.
POST --> http://localhost:8000/registertestmethod/
{
"name": "Name",
"url": "https://www.google.com"
}
returns
{
"tokenid": 32 #unique token ID return
}
This tokenid should be long I am using id as of now.
GET --> http://localhost:8000/checkstatus/32
is returning
{
"tokenid": 32, #unique tokenid refering to register info
"Runtestnow": false, #if job is running
"VMcount": 0 # number of VM used in sauce lab by the Jobs
}
DELETE --> http://localhost:8000/releasetoken/32 #should delete the jobs after done.
is deleting and returning
{
"Deleted": true
}
I want it to be dynamic and store info in database. Token should return everything on checkstatus.
You can use the ModelViewSet approach for this, this here is a very simple example for an API endpoint.
views.py
from rest_framework.viewsets import ModelViewSet
from api.serializers import SaucelabsSerializer
from rest_framework.response import Response
class SaucelabModelViewSet(ModelViewSet):
serializer_class = SaucelabSerializer
queryset = Sauselab.objects.all()
http_method_names = ['get', 'head', 'options', 'post']
def create(self, request):
pnam = request.data.get('project_name', None)
url = request.data.get('url', None)
if pnam and url:
# do something here
return Response({'success': 'Your success message'}, status=status.HTTP_200_OK)
else:
return Response({"error": "Your error message"}, status=status.HTTP_400_BAD_REQUEST)
serializers.py
from rest_framework.serializer import ModelSerializer
from appname.models import Saucelab
class SaucelabSerializer(ModelSerializer):
class Meta:
model = Saucelab
fields = '__all__'
appname/models.py
from django.db import models
class Saucelab(models.Model)
project_name = models.CharField(max_length=255)
url = models.URLField()
urls.py
from rest_framework.routers import DefaultRouter
from api import views
router = DefaultRouter()
router.register('your-endpoint-name', views.SaucelabModelViewSet, basename='your-url-name')
urlpatterns = []
urlpatterns += router.urls
This is a very basic example, where you create a model called saucelab which has the two fields you require, ie.project_name and url.
We create a app called api with two files inside it that aren't auto generated, serializers.py and urls.py. We create the most basic model serializer and ask it to serialize all fields of model Saucelab. Then we create a simple modelviewset which out of the box gives you a CRUD functionality. You can override the create method if you need to run some specific conditions otherwise dont override any methods and just make request to the endpoint with appropriate HTTP methods.
Here are a few links you can read
https://www.django-rest-framework.org/api-guide/viewsets/#modelviewset
https://www.django-rest-framework.org/api-guide/serializers/#modelserializer
To generate a random token
from django.utils.crypto import get_random_string
print(get_random_string(length=25))
output
u'rRXVe68NO7m3mHoBS488KdHaqQPD6Ofv'

How to limit the format of django-rest-framework-api uploading(post method)

In fact, I need your help to limit my API to upload
I only need Jason and excel formats(csv,xls,xlsl) on the server-side, but I did not find any code for that.
The second question is, what methods should I use to authenticate the user?
Here's a link that I use its code in my API
in fact, I need fix this part:
from django.shortcuts import render
# Create your views here.
from rest_framework.views import APIView
from rest_framework.parsers import MultiPartParser, FormParser, JSONParser
from rest_framework.response import Response
from rest_framework import status
from .serializers import FileSerializer
class FileView(APIView):
parser_classes = (MultiPartParser, FormParser)
def post(self, request, *args, **kwargs):
file_serializer = FileSerializer(data=request.data)
if file_serializer.is_valid():
file_serializer.save()
return Response(file_serializer.data, status=status.HTTP_201_CREATED)
else:
return Response(file_serializer.errors, status=status.HTTP_400_BAD_REQUEST)
def get():
pass
def update():
pass
my model :
from django.db import models
class File(models.Model):
file = models.FileField(blank=False, null=False)
remark = models.CharField(max_length=20)
timestamp = models.DateTimeField(auto_now_add=True)
and my serializers :
from rest_framework import serializers
from .models import File
class FileSerializer(serializers.ModelSerializer):
class Meta():
model = File
fields = ('file', 'remark', 'timestamp')
but I didn't know how most make it limit
my Django version is 2 AND my python version is 3
I will thank you if you help me :)
have to go coding time

Upload image file in Django REST framework with angularJS

I'm currently uploading images through an API created by with Django REST framework with angularJS.
However, the problem is that it is stored as a media file but not stored in a database.
Here is my ViewSet code.
# views.py
import time
from rest_framework import viewsets
from rest_framework.permissions import IsAdminUser
from django.contrib.auth import get_user_model
from .serializers import UserSerializer
class UserViewSet(viewsets.ModelViewSet):
permission_classes = (IsAdminUser,)
serializer_class = UserSerializer
queryset = get_user_model()
def perform_update(self, serializer):
serializer.save(profile_image=self.request.data.get('file'))
But, when I added the code time.sleep(1) before serializer.save() function, it was mysteriously saved to the database successfully.
def perform_update(self, serializer):
time.sleep(1) # Delay 1s
serializer.save(profile_image=self.request.data.get('file'))
Why does this happen? And please give me a wise solution.
I don't know if it would help, so I attached the angularjs code.
.service('fileUpload', ['$http', function ($http) {
this.uploadFileToUrl = function(file, uploadUrl){
var fd = new FormData();
fd.append('file', file);
return $http.patch(uploadUrl, fd, {
transformRequest: angular.identity,
headers: {'Content-Type': undefined}
})
}
}])
This is my User model I was requested to share. I omitted the code that are not related to the error.
I confirmed that the image was saved in the path I specified. But not saved in database.
# models.py
import datetime
import os
from django.contrib.auth.models import AbstractBaseUser, PermissionsMixin
from django.db import models
def set_filename_format(now, instance, filename):
return "{username}-{date}-{microsecond}{extension}".format(
username=instance.username,
date=now.date().strftime("%Y-%m-%d"),
microsecond=now.microsecond,
extension=os.path.splitext(filename)[1],
)
def profile_image_path(instance, filename):
now = datetime.datetime.now()
path = "profile_images/{username}/{year}/{month}/{day}/{filename}".format(
username=instance.username,
year=now.year,
month=now.month,
day=now.day,
filename=set_filename_format(now, instance, filename),
)
return path
class User(AbstractBaseUser, PermissionsMixin):
# ...
profile_image = models.ImageField(
upload_to=profile_image_path,
blank=True,
)
# ...

Django Rest Framework File Upload

I am using Django Rest Framework and AngularJs to upload a file. My view file looks like this:
class ProductList(APIView):
authentication_classes = (authentication.TokenAuthentication,)
def get(self,request):
if request.user.is_authenticated():
userCompanyId = request.user.get_profile().companyId
products = Product.objects.filter(company = userCompanyId)
serializer = ProductSerializer(products,many=True)
return Response(serializer.data)
def post(self,request):
serializer = ProductSerializer(data=request.DATA, files=request.FILES)
if serializer.is_valid():
serializer.save()
return Response(data=request.DATA)
As the last line of post method should return all the data, I have several questions:
how to check if there is anything in request.FILES?
how to serialize file field?
how should I use parser?
Editor's note:
This answer uses pre_save, which no longer exists in Django REST framework 3.0.
In a sufficiently new version of Django REST framework, MultiPartParser should be available by default, which allows uploading file with no special handling. See an answer below for an example.
I'm using the same stack and was also looking for an example of file upload, but my case is simpler since I use the ModelViewSet instead of APIView. The key turned out to be the pre_save hook. I ended up using it together with the angular-file-upload module like so:
# Django
class ExperimentViewSet(ModelViewSet):
queryset = Experiment.objects.all()
serializer_class = ExperimentSerializer
def pre_save(self, obj):
obj.samplesheet = self.request.FILES.get('file')
class Experiment(Model):
notes = TextField(blank=True)
samplesheet = FileField(blank=True, default='')
user = ForeignKey(User, related_name='experiments')
class ExperimentSerializer(ModelSerializer):
class Meta:
model = Experiment
fields = ('id', 'notes', 'samplesheet', 'user')
// AngularJS
controller('UploadExperimentCtrl', function($scope, $upload) {
$scope.submit = function(files, exp) {
$upload.upload({
url: '/api/experiments/' + exp.id + '/',
method: 'PUT',
data: {user: exp.user.id},
file: files[0]
});
};
});
Use the FileUploadParser, it's all in the request.
Use a put method instead, you'll find an example in the docs :)
class FileUploadView(views.APIView):
parser_classes = (FileUploadParser,)
def put(self, request, filename, format=None):
file_obj = request.FILES['file']
# do some stuff with uploaded file
return Response(status=204)
Finally I am able to upload image using Django. Here is my working code
views.py
class FileUploadView(APIView):
parser_classes = (FileUploadParser, )
def post(self, request, format='jpg'):
up_file = request.FILES['file']
destination = open('/Users/Username/' + up_file.name, 'wb+')
for chunk in up_file.chunks():
destination.write(chunk)
destination.close() # File should be closed only after all chuns are added
# ...
# do some stuff with uploaded file
# ...
return Response(up_file.name, status.HTTP_201_CREATED)
urls.py
urlpatterns = patterns('',
url(r'^imageUpload', views.FileUploadView.as_view())
curl request to upload
curl -X POST -S -H -u "admin:password" -F "file=#img.jpg;type=image/jpg" 127.0.0.1:8000/resourceurl/imageUpload
From my experience, you don't need to do anything particular about file fields, you just tell it to make use of the file field:
from rest_framework import routers, serializers, viewsets
class Photo(django.db.models.Model):
file = django.db.models.ImageField()
def __str__(self):
return self.file.name
class PhotoSerializer(serializers.ModelSerializer):
class Meta:
model = models.Photo
fields = ('id', 'file') # <-- HERE
class PhotoViewSet(viewsets.ModelViewSet):
queryset = models.Photo.objects.all()
serializer_class = PhotoSerializer
router = routers.DefaultRouter()
router.register(r'photos', PhotoViewSet)
api_urlpatterns = ([
url('', include(router.urls)),
], 'api')
urlpatterns += [
url(r'^api/', include(api_urlpatterns)),
]
and you're ready to upload files:
curl -sS http://example.com/api/photos/ -F 'file=#/path/to/file'
Add -F field=value for each extra field your model has. And don't forget to add authentication.
After spending 1 day on this, I figured out that ...
For someone who needs to upload a file and send some data, there is no straight fwd way you can get it to work. There is an open issue in JSON API specs for this. One possibility I have seen is to use multipart/related as shown here, but I think it's very hard to implement in DRF.
Finally what I implemented was to send the request as FormData. You would send each file as file and all other data as text.
Now for sending the data as text you have two choices. case 1) you can send each data as a key-value pair or case 2) you can have a single key called data and send the whole JSON as a string in value.
The first method would work out of the box if you have simple fields but it will be an issue if you have nested serializes. The multipart parser won't be able to parse the nested fields.
Below I am providing the implementation for both the cases
models.py
class Posts(models.Model):
id = models.UUIDField(default=uuid.uuid4, primary_key=True, editable=False)
caption = models.TextField(max_length=1000)
media = models.ImageField(blank=True, default="", upload_to="posts/")
tags = models.ManyToManyField('Tags', related_name='posts')
serializers.py -> no special changes needed, not showing my serializer here as it's too lengthy because of the writable ManyToMany Field implementation.
views.py
class PostsViewset(viewsets.ModelViewSet):
serializer_class = PostsSerializer
#parser_classes = (MultipartJsonParser, parsers.JSONParser) use this if you have simple key value pair as data with no nested serializers
#parser_classes = (parsers.MultipartParser, parsers.JSONParser) use this if you want to parse json in the key value pair data sent
queryset = Posts.objects.all()
lookup_field = 'id'
Now, if you are following the first method and are only sending non-Json data as key-value pairs, you don't need a custom parser class. DRF'd MultipartParser will do the job. But for the second case or if you have nested serializers (like I have shown) you will need a custom parser as shown below.
utils.py
from django.http import QueryDict
import json
from rest_framework import parsers
class MultipartJsonParser(parsers.MultiPartParser):
def parse(self, stream, media_type=None, parser_context=None):
result = super().parse(
stream,
media_type=media_type,
parser_context=parser_context
)
data = {}
# for case1 with nested serializers
# parse each field with json
for key, value in result.data.items():
if type(value) != str:
data[key] = value
continue
if '{' in value or "[" in value:
try:
data[key] = json.loads(value)
except ValueError:
data[key] = value
else:
data[key] = value
# for case 2
# find the data field and parse it
data = json.loads(result.data["data"])
qdict = QueryDict('', mutable=True)
qdict.update(data)
return parsers.DataAndFiles(qdict, result.files)
This serializer would basically parse any JSON content in the values.
The request example in postman for both cases:
Case 1
Case 2
If anyone interested in the easiest example with ModelViewset for Django Rest Framework.
The Model is,
class MyModel(models.Model):
name = models.CharField(db_column='name', max_length=200, blank=False, null=False, unique=True)
imageUrl = models.FileField(db_column='image_url', blank=True, null=True, upload_to='images/')
class Meta:
managed = True
db_table = 'MyModel'
The Serializer,
class MyModelSerializer(serializers.ModelSerializer):
class Meta:
model = MyModel
fields = "__all__"
And the View is,
class MyModelView(viewsets.ModelViewSet):
queryset = MyModel.objects.all()
serializer_class = MyModelSerializer
Test in Postman,
models.py
from django.db import models
import uuid
class File(models.Model):
id = models.UUIDField(primary_key=True, default=uuid.uuid4, editable=False)
file = models.FileField(blank=False, null=False)
def __str__(self):
return self.file.name
serializers.py
from rest_framework import serializers
from .models import File
class FileSerializer(serializers.ModelSerializer):
class Meta:
model = File
fields = "__all__"
views.py
from django.shortcuts import render
from rest_framework.parsers import FileUploadParser
from rest_framework.response import Response
from rest_framework.views import APIView
from rest_framework import status
from .serializers import FileSerializer
class FileUploadView(APIView):
permission_classes = []
parser_class = (FileUploadParser,)
def post(self, request, *args, **kwargs):
file_serializer = FileSerializer(data=request.data)
if file_serializer.is_valid():
file_serializer.save()
return Response(file_serializer.data, status=status.HTTP_201_CREATED)
else:
return Response(file_serializer.errors, status=status.HTTP_400_BAD_REQUEST)
urls.py
from apps.files import views as FileViews
urlpatterns = [
path('api/files', FileViews.FileUploadView.as_view()),
]
settings.py
# file uload parameters
MEDIA_URL = '/media/'
MEDIA_ROOT = os.path.join(BASE_DIR, 'media')
Send a post request to api/files with a your file attached to a form-data field file. The file will be uploaded to /media folder and a db record will be added with id and file name.
I solved this problem with ModelViewSet and ModelSerializer. Hope this will help community.
I also preffer to have validation and Object->JSON (and vice-versa) login in serializer itself rather than in views.
Lets understand it by example.
Say, I want to create FileUploader API. Where it will be storing fields like id, file_path, file_name, size, owner etc in database. See sample model below:
class FileUploader(models.Model):
file = models.FileField()
name = models.CharField(max_length=100) #name is filename without extension
version = models.IntegerField(default=0)
upload_date = models.DateTimeField(auto_now=True, db_index=True)
owner = models.ForeignKey('auth.User', related_name='uploaded_files')
size = models.IntegerField(default=0)
Now, For APIs this is what I want:
GET:
When I fire the GET endpoint, I want all above fields for every uploaded file.
POST:
But for user to create/upload file, why she has to worry about passing all these fields. She can just upload the file and then, I suppose, serializer can get rest of the fields from uploaded FILE.
Searilizer:
Question: I created below serializer to serve my purpose. But not sure if its the right way to implement it.
class FileUploaderSerializer(serializers.ModelSerializer):
# overwrite = serializers.BooleanField()
class Meta:
model = FileUploader
fields = ('file','name','version','upload_date', 'size')
read_only_fields = ('name','version','owner','upload_date', 'size')
def validate(self, validated_data):
validated_data['owner'] = self.context['request'].user
validated_data['name'] = os.path.splitext(validated_data['file'].name)[0]
validated_data['size'] = validated_data['file'].size
#other validation logic
return validated_data
def create(self, validated_data):
return FileUploader.objects.create(**validated_data)
Viewset for reference:
class FileUploaderViewSet(viewsets.ModelViewSet):
serializer_class = FileUploaderSerializer
parser_classes = (MultiPartParser, FormParser,)
# overriding default query set
queryset = LayerFile.objects.all()
def get_queryset(self, *args, **kwargs):
qs = super(FileUploaderViewSet, self).get_queryset(*args, **kwargs)
qs = qs.filter(owner=self.request.user)
return qs
I'd like to write another option that I feel is cleaner and easier to maintain. We'll be using the defaultRouter to add CRUD urls for our viewset and we'll add one more fixed url specifying the uploader view within the same viewset.
**** views.py
from rest_framework import viewsets, serializers
from rest_framework.decorators import action, parser_classes
from rest_framework.parsers import JSONParser, MultiPartParser
from rest_framework.response import Response
from rest_framework_csv.parsers import CSVParser
from posts.models import Post
from posts.serializers import PostSerializer
class PostsViewSet(viewsets.ModelViewSet):
queryset = Post.objects.all()
serializer_class = PostSerializer
parser_classes = (JSONParser, MultiPartParser, CSVParser)
#action(detail=False, methods=['put'], name='Uploader View', parser_classes=[CSVParser],)
def uploader(self, request, filename, format=None):
# Parsed data will be returned within the request object by accessing 'data' attr
_data = request.data
return Response(status=204)
Project's main urls.py
**** urls.py
from rest_framework import routers
from posts.views import PostsViewSet
router = routers.DefaultRouter()
router.register(r'posts', PostsViewSet)
urlpatterns = [
url(r'^posts/uploader/(?P<filename>[^/]+)$', PostsViewSet.as_view({'put': 'uploader'}), name='posts_uploader')
url(r'^', include(router.urls), name='root-api'),
url('admin/', admin.site.urls),
]
.- README.
The magic happens when we add #action decorator to our class method 'uploader'. By specifying "methods=['put']" argument, we are only allowing PUT requests; perfect for file uploading.
I also added the argument "parser_classes" to show you can select the parser that will parse your content. I added CSVParser from the rest_framework_csv package, to demonstrate how we can accept only certain type of files if this functionality is required, in my case I'm only accepting "Content-Type: text/csv".
Note: If you're adding custom Parsers, you'll need to specify them in parsers_classes in the ViewSet due the request will compare the allowed media_type with main (class) parsers before accessing the uploader method parsers.
Now we need to tell Django how to go to this method and where can be implemented in our urls. That's when we add the fixed url (Simple purposes). This Url will take a "filename" argument that will be passed in the method later on. We need to pass this method "uploader", specifying the http protocol ('PUT') in a list to the PostsViewSet.as_view method.
When we land in the following url
http://example.com/posts/uploader/
it will expect a PUT request with headers specifying "Content-Type" and Content-Disposition: attachment; filename="something.csv".
curl -v -u user:pass http://example.com/posts/uploader/ --upload-file ./something.csv --header "Content-type:text/csv"
Some of the solutions are deprecated (request.data should be used for Django 3.0+). Some of them do not validate the input. Also, I would appreciate a solution with swagger annotation. So I recommend using the following code:
from drf_yasg.utils import swagger_auto_schema
from rest_framework import serializers
from rest_framework.parsers import MultiPartParser
from rest_framework.response import Response
from rest_framework.views import APIView
class FileUploadAPI(APIView):
parser_classes = (MultiPartParser, )
class InputSerializer(serializers.Serializer):
image = serializers.ImageField()
#swagger_auto_schema(
request_body=InputSerializer
)
def put(self, request):
input_serializer = self.InputSerializer(data=request.data)
input_serializer.is_valid(raise_exception=True)
# process file
file = input_serializer.validated_data['image']
return Response(status=204)
I have used this view to upload file to aws. Here upload_file is a helper function while overall you can use this view to get upload the file in form-data.
class FileUploadView(GenericAPIView):
def post(self, request):
try:
file = request.data['file']
if file.content_type == 'image/png' or file.content_type == 'image/jpeg':
file_name = upload_file(file)
return Response({"name": file_name}, status=status.HTTP_202_ACCEPTED)
else:
raise UnsupportedMediaType(file.content_type)
except KeyError:
return Response("file missing.", status=status.HTTP_404_NOT_FOUND)
Best Straightforward Way to handle single upload file or multiple files in a single request is this
#api_view(['POST'])
def file_list(request): # use APIview or function based view or any view u want
# for single file
file = request.FILES["file"]
print(file)
# Do what ever you want with it
# for multiple file
files = request.FILES.getlist('file')
for file in files:
print(file)
# Do what ever you want with it
from rest_framework import status
from rest_framework.response import Response
class FileUpload(APIView):
def put(request):
try:
file = request.FILES['filename']
#now upload to s3 bucket or your media file
except Exception as e:
print e
return Response(status,
status.HTTP_500_INTERNAL_SERVER_ERROR)
return Response(status, status.HTTP_200_OK)
If you are using ModelViewSet, well actually you are done! It handles every things for you! You just need to put the field in your ModelSerializer and set content-type=multipart/form-data; in your client.
BUT as you know you can not send files in json format. (when content-type is set to application/json in your client). Unless you use Base64 format.
So you have two choices:
let ModelViewSet and ModelSerializer handle the job and send the request using content-type=multipart/form-data;
set the field in ModelSerializer as Base64ImageField (or) Base64FileField and tell your client to encode the file to Base64 and set the content-type=application/json
from rest_framework import status, generics
from rest_framework.response import Response
from rest_framework import serializers
import logging
logger = logging.getLogger(__name__)`enter code here`
class ImageUploadSerializer(serializers.Serializer):
file = serializers.FileField()
class UploadImages(generics.GenericAPIView):
serializer_class = ImageUploadSerializer
permission_classes = [IsAuthenticated, ]
def post(self, request):
try:
data = self.serializer_class(data=request.data)
if data.is_valid() is False:
return Response({'error': ERROR_MESSAGES.get('400')}, status=status.HTTP_400_BAD_REQUEST)
is_file_upload_success, file_item = save_aws_article_image(data.validated_data.get('file'),
request.user, upload_type)
if is_file_upload_success:
logger.info('{0} file uploaded {1}'.format(file_item['file_obj'].path, datetime.now()))
return Response({'path': file_item['file_obj'].path, 'id': file_item['file_obj'].uuid,
'name': file_item['file_obj'].name},
status=status.HTTP_201_CREATED)
except Exception as e:
logger.error(e, exc_info=True)
return Response({"error": e}, status=status.HTTP_500_INTERNAL_SERVER_ERROR)
In django-rest-framework request data is parsed by the Parsers.
http://www.django-rest-framework.org/api-guide/parsers/
By default django-rest-framework takes parser class JSONParser. It will parse the data into json. so, files will not be parsed with it.
If we want files to be parsed along with other data we should use one of the below parser classes.
FormParser
MultiPartParser
FileUploadParser
def post(self,request):
serializer = ProductSerializer(data=request.DATA, files=request.FILES)
if serializer.is_valid():
serializer.save()
return Response(serializer.data)
This is the one of the approach I've applied hopefully it'll help.
class Model_File_update(APIView):
parser_classes = (MultiPartParser, FormParser)
permission_classes = [IsAuthenticated] # it will check if the user is authenticated or not
authentication_classes = [JSONWebTokenAuthentication] # it will authenticate the person by JSON web token
def put(self, request):
id = request.GET.get('id')
obj = Model.objects.get(id=id)
serializer = Model_Upload_Serializer(obj, data=request.data)
if serializer.is_valid():
serializer.save()
return Response(serializer.data, status=200)
else:
return Response(serializer.errors, status=400)
You can generalize #Nithin's answer to work directly with DRF's existing serializer system by generating a parser class to parse specific fields which are then fed directly into the standard DRF serializers:
from django.http import QueryDict
import json
from rest_framework import parsers
def gen_MultipartJsonParser(json_fields):
class MultipartJsonParser(parsers.MultiPartParser):
def parse(self, stream, media_type=None, parser_context=None):
result = super().parse(
stream,
media_type=media_type,
parser_context=parser_context
)
data = {}
# find the data field and parse it
qdict = QueryDict('', mutable=True)
for json_field in json_fields:
json_data = result.data.get(json_field, None)
if not json_data:
continue
data = json.loads(json_data)
if type(data) == list:
for d in data:
qdict.update({json_field: d})
else:
qdict.update({json_field: data})
return parsers.DataAndFiles(qdict, result.files)
return MultipartJsonParser
This is used like:
class MyFileViewSet(ModelViewSet):
parser_classes = [gen_MultipartJsonParser(['tags', 'permissions'])]
# ^^^^^^^^^^^^^^^^^^^
# Fields that need to be further JSON parsed
....
A DRF viewset fileupload example with React(axios) to send an audioBlob:
class MyViewSet(viewsets.ModelViewSet):
parser_classes = (MultiPartParser, FormParser)
queryset = MyModel.objects.all().order_by('created_at')
serializer_class = MySerializer
serializer:
class MySerializer(serializers.ModelSerializer):
class Meta:
model = MyModel
fields = '__all__'
model:
class MyModel(models.Model):
sentence = models.ForeignKey(Sentence, related_name="voice_sentence", on_delete=models.CASCADE)
voice_record = models.FileField(blank=True, default='')
created_at = models.DateTimeField(auto_now_add=True)
axios:
export const sendSpeechText = async (audioBlob: any) => {
const headers = {
'Content-Type': 'application/json',
'Content-Disposition': 'attachment; filename=audiofile.webm'
}
const audiofile = new File([audioBlob], "audiofile.webm", { type: "audio/webm" })
const formData = new FormData();
formData.append("sentence", '1');
formData.append("voice_record", audiofile);
return await axios.post(
SEND_SPEECH_URL,
formData,
{
crossDomain: true,
headers: headers
},
)
}
NOTE: voice_record in formData should be the same in your model
There are majorly 3 ways for upload files in drf
suppose you have Tag model with title and logo fields and TagSerializer
class Tag(models.Model):
title = models.CharField(max_length=10, default='')
file = models.FileField(upload_to='tag/', blank=True, null=True, )
class TagSerializer(rest_serializers.ModelSerializer):
class Meta:
model = Tag
fields = '__all__'
you can choose one of them according to your situation.
1- using serializer:
class UploadFile(APIView):
parser_classes = (MultiPartParser, FormParser)
def post(self, request):
serializer = TagSerializer(data=request.data)
serializer.is_valid(raise_exception=True)
serializer.save()
return Response(serializer.data, status=status.HTTP_200_OK)
2- using write method :
def save_file(file: InMemoryUploadedFile, full_path):
with open(full_path, 'wb+') as f:
for chunk in file.chunks():
f.write(chunk)
class UploadFile(APIView):
parser_classes = (MultiPartParser, FormParser)
def post(self, request):
file: InMemoryUploadedFile = request.FILES['file']
# define file_save_path variable
full_path = file_save_path + file.name
save_file(file, full_path)
return Response(serializer.data, status=status.HTTP_200_OK)
3- using FileSystemStorage:
class UploadFile(APIView):
parser_classes = (MultiPartParser, FormParser)
def post(self, request):
file: InMemoryUploadedFile = request.FILES['file']
f = FileSystemStorage()
# this will save file in MEDIA_ROOT path
f.save(file.name, file)
return Response(serializer.data, status=status.HTTP_200_OK)
For the users who want to use or prefer a Function-Based Views for uploading files.
This is a Complete Guide from Creating Models > views > Serializers > URLs and Testing the endpoint with Postman. I have put the comments inside the code where required.
# models.py
# Imports
from django.db import models
import os
def document_path_and_name(instance, filename):
''' Change the filename to 'instance_id + document_name '''
ext = filename.split('.')[-1]
filename = "%s_%s.%s" % (instance.id, instance.document_name, ext)
''' if document_name is 'doucment one' in pdf and id is 1
then filname will be saved as = 1_document_one.pdf '''
return os.path.join('files/', filename)
class Document(models.Model):
# I'm using document_name and id to give the filename that would be save with
# this using document_path_and_name function.
# you can modify on your need.
document_name = models.CharField(max_length=100)
file = models.FileField(upload_to=document_path_and_name)
def __str__(self):
return self.document_name
We don't need a Serializer to validate the file upload here but would need one if we need to serialize the response. So let's go with a simple ReadOnly Serializer in this case.
# serializers.py
# imports
from rest_framework import serializers
class DocumentSerializer(serializers.Serializer):
id = serializers.IntegerField(read_only=True)
document_name = serializers.CharField(read_only=True)
file = serializers.URLField(read_only=True)
Now in the api_view, we will be using the MultiPartParser decorator to upload files via a POST request. We would need a document_name and a file for this function to upload the file correctly as we had set the Model.
# views.py
# imports
from rest_framework.decorators import api_view, parser_classes
from rest_framework.response import Response
from rest_framework.parsers import MultiPartParser
from .models import Document
from .serializers import DocumentSerializer
#api_view(['POST'])
#parser_classes([MultiPartParser])
def upload_document(request, filename, format=None):
"""
A view that can accept POST requests with .media_type: multipart/form-data content.
"""
file = request.FILES['file']
doc = Document.objects.create(document_name=filename, file=file)
# Do any thing else here
serializer = DocumentSerializer(doc, many=False)
return Response(serializer.data)
So, We will be passing document_name in the URL param, we can call it anything but I defined it as the filename.
and our API ENDPOINT or Url will be like;
# imports
from django.urls import path
from .views import upload_document
urlpatterns = [
path('upload_document/<str:filename>/', upload_document),
]
So to test this via Postman, go to your valid API endpoint like below
I'm passing the filename for the document_name you can pass anything. You would notice that the actual file name is something else in pdf format in the screenshot below. That will be replaced with the help of our document_path_and_name function to id_document_name. So here the save filename is 1_filename.pdf
Now just make a request and your file will be uploaded to your directed file storage path. And you will get the JSON Response from DocumentSerializer.
The main thing which was responsible for the file upload is the MultiPartParser decorator. Must visit the Docs for more details.
If you're using ViewSets, you can add a custom action to handle file uploads:
from rest_framework.decorators import action
from rest_framework.response import Response
from rest_framework.parsers import FileUploadParser
from rest_framework import viewsets
...
class SomeViewSet(viewsets.ModelViewSet):
serializer_class = ...
permission_classes = [...]
queryset = ...
#action(methods=['put'], detail=True, parser_classes=[FileUploadParser])
def upload_file(self, request, pk=None):
obj = self.get_object()
obj.file = request.data['file']
obj.save()
return Response(status=204)
This keeps everything within the ViewSet. You'll get an endpoint that looks something like this api/item/32/upload_file/.
The reason you'd use FileUploadParser as opposed to other options like multipart is if you're uploading from a native app for example and don't want to rely on a multi part encoder.

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