FileUploadParser doesn't get the file name

问题

I just want to create a REST API that receives a file, process it and return some information. My problem is that I am following this example: http://www.django-rest-framework.org/api-guide/parsers/#fileuploadparser

And I can't make it work using Postman or curl, I think I am missing something. The parser always gives me these two errors:

FileUpload parse error - none of upload handlers can handle the stream
Missing filename. Request should include a Content-Disposition header with a filename parameter.

This is the code:

views.py:

class FileUploadView(APIView):
    parser_classes = (FileUploadParser,)

    def post(self, request, filename, format=None):
        file_obj = request.data['file']
        # ...
        # do some stuff with uploaded file
        # ...
        return Response(status=204)

    def put(self, request, filename, format=None):
        file_obj = request.data['file']
        # ...
        # do some stuff with uploaded file
        # ...
        return Response(status=204)

urls.py

urlpatterns = [
   url(r'predict/(?P<filename>[^/]+)$', app.views.FileUploadView.as_view())
]

settings.py

"""
Django settings for GenderAPI project.

Generated by 'django-admin startproject' using Django 1.9.1.

For more information on this file, see
https://docs.djangoproject.com/en/1.9/topics/settings/

For the full list of settings and their values, see
https://docs.djangoproject.com/en/1.9/ref/settings/
"""

import os
import posixpath


LOGGING = {
    'version': 1,
    'disable_existing_loggers': False,
    'handlers': {
        'file': {
            'level': 'DEBUG',
            'class': 'logging.FileHandler',
            'filename': 'debug.log',
        },
    },
    'loggers': {
        'django': {
            'handlers': ['file'],
            'level': 'DEBUG',
            'propagate': True,
        },
    },
}

# Build paths inside the project like this: os.path.join(BASE_DIR, ...)
BASE_DIR = os.path.dirname(os.path.dirname(os.path.abspath(__file__)))


# Quick-start development settings - unsuitable for production
# See https://docs.djangoproject.com/en/1.9/howto/deployment/checklist/

# SECURITY WARNING: keep the secret key used in production secret!
SECRET_KEY = removed

# SECURITY WARNING: don't run with debug turned on in production!
DEBUG = True

ALLOWED_HOSTS = ['localhost','127.0.0.1']

REST_FRAMEWORK = {
    'DEFAULT_PARSER_CLASSES': (
        'rest_framework.parsers.FileUploadParser'
    )
}


# Application definition

INSTALLED_APPS = [    

    # Add your apps here to enable them
    'django.contrib.admin',
    'django.contrib.auth',
    'django.contrib.contenttypes',
    'django.contrib.sessions',
    'django.contrib.messages',
    'django.contrib.staticfiles',   
    'rest_framework',
    'app'  


]

MIDDLEWARE = [

    'django.middleware.security.SecurityMiddleware',
    'django.contrib.sessions.middleware.SessionMiddleware',
    'django.middleware.common.CommonMiddleware',
    'django.middleware.csrf.CsrfViewMiddleware',
    'django.contrib.auth.middleware.AuthenticationMiddleware',
    'django.contrib.auth.middleware.SessionAuthenticationMiddleware',
    'django.contrib.messages.middleware.MessageMiddleware',
    'django.middleware.clickjacking.XFrameOptionsMiddleware'
]

ROOT_URLCONF = 'GenderAPI.urls'

TEMPLATES = [
    {
        'BACKEND': 'django.template.backends.django.DjangoTemplates',
        'DIRS': [],
        'APP_DIRS': True,
        'OPTIONS': {
            'context_processors': [
                'django.template.context_processors.debug',
                'django.template.context_processors.request',
                'django.contrib.auth.context_processors.auth',
                'django.contrib.messages.context_processors.messages',
            ],
        },
    },
]

WSGI_APPLICATION = 'GenderAPI.wsgi.application'

# Database
# https://docs.djangoproject.com/en/1.9/ref/settings/#databases

DATABASES = {
    'default': {
        'ENGINE': 'django.db.backends.sqlite3',
        'NAME': os.path.join(BASE_DIR, 'db.sqlite3'),
    }
}

# Static files (CSS, JavaScript, Images)
# https://docs.djangoproject.com/en/1.9/howto/static-files/

STATIC_URL = '/static/'
STATIC_ROOT = posixpath.join(*(BASE_DIR.split(os.path.sep) + ['static']))

FILE_UPLOAD_TEMP_DIR = BASE_DIR
MEDIA_URL  = '/media/'

Here you can see a postman capture (I have tried everything):

PUT /predict/pabloGrande.jpg HTTP/1.1
Host: 127.0.0.1:52276
Content-Type: multipart/form-data; boundary=----WebKitFormBoundary7MA4YWxkTrZu0gW

------WebKitFormBoundary7MA4YWxkTrZu0gW
Content-Disposition: form-data; name="file"; filename="04320cf.jpg"
Content-Type: image/jpeg


------WebKitFormBoundary7MA4YWxkTrZu0gW--

requirements:

bleach==1.5.0
Django==1.11.6
djangorestframework==3.7.1
html5lib==0.9999999
Markdown==2.6.9
numpy==1.13.3
olefile==0.44
pandas==0.20.3
Pillow==4.3.0
pip==9.0.1
protobuf==3.4.0
python-dateutil==2.6.1
pytz==2017.2
scipy==1.0.0rc1
setuptools==28.8.0
six==1.11.0
tensorflow==1.3.0
tensorflow-tensorboard==0.1.8
Werkzeug==0.12.2
wheel==0.30.0

Thanks so much for your help

回答1:

In django REST framework. we have components like Parsers, Renderers and Serializers.

The responsibility of Parsers is to parse the data that is sent by request methods GET, POST and PUT, etc.
Default parser used in django REST is 'JSONParser'. It only parses the data JSON data[numbers, string, date]. It ignores the data like FILES.

In order to parse the FILES we need to use parsers like "MultiPartParser" or "FormParser".

Example Code :

    from rest_framework.parsers import MultiPartParser
    from rest_framework.response import Response
    from rest_framework.views import APIView

    class ExampleView(APIView):
        """
        A view that can accept POST requests with JSON content.
        """
        parser_classes = (MultiPartParser,)

        def post(self, request, format=None):
            # to access files
            print request.FILES
            # to access data
            print request.data
            return Response({'received data': request.data})

When we use property request.data then parser will parse the data.

References: Django REST Docs, Django REST Github

回答2:

I face the same problem.The problem error message shows:

{"detail":"Missing filename. Request should include a Content-Disposition header with a filename parameter."} I do all the steps above my answer but it doesn't work.Finally,

I find the reason is in the backend of viewset.

it shows like this

parser_classes = (FileUploadParser, MultiPartParser, FormParser)

and I remove the FileUploadParser

  parser_classes = ( MultiPartParser, FormParser)

and It works, so I think you should pay more attention to it

回答3:

in your views.py change the parser like this

parser_classes = (JSONParser, MultiPartParser)

回答4:

On the second error Missing filename. Request should include a Content-Disposition header with a filename parameter. using postman I removed the header Content-Type :multipart/form-data and was successful.

It seems the issue is the custom Content-Type header overrides the default Content-Type header that should be sent. Check this thread for reference Content-Type for multipart posts

回答5:

You don't need to use MultipartParser or FormParser.

What you need is a serializer with a FileField() like so:

serializers.py:

class FileUploadSerializer(serializers.Serializer):
    # I set use_url to False so I don't need to pass file 
    # through the url itself - defaults to True if you need it
    file = serializers.FileField(use_url=False)

So then when you try to access file here below, you'll have a dict with a key named file. Personally, I'd probably name it something more descriptive than just "file", but that's up to you.

from .serializers import FileUploadSerializer

    class FileUploadView(APIView):

       def post(self, request):
           # set 'data' so that you can use 'is_vaid()' and raise exception
           # if the file fails validation
           serializer = FileUploadSerializer(data=request.data)
           serializer.is_valid(raise_exception=True)
           # once validated, grab the file from the request itself
           file = request.FILES['file']

来源：https://stackoverflow.com/questions/46806335/fileuploadparser-doesnt-get-the-file-name

标签

django

django-rest-framework