django - FileUploadParser 没有得到文件名

Question

我只想创建一个接收文件、处理文件并返回一些信息的 REST API。我的问题是我正在遵循这个例子:
http://www.django-rest-framework.org/api-guide/parsers/#fileuploadparser

而且我无法使用 Postman 或 curl 使其工作，我想我错过了一些东西。解析器总是给我这两个错误:

FileUpload 解析错误 - 没有任何上传处理程序可以处理流

缺少文件名。请求应包含带有文件名参数的 Content-Disposition header 。

这是代码:

View .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)

网址.py

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

设置.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/'

在这里你可以看到 postman 捕获(我已经尝试了一切):

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--

要求:

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

非常感谢你的帮助

Answer 1

您不需要使用 MultipartParser或 FormParser .

您需要的是带有 FileField() 的序列化程序像这样:

序列化程序.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)

那么当你尝试访问 file在下面，您将拥有一个带有名为 file 的键的 dict .就个人而言，我可能会将其命名为比"file"更具描述性的名称，但这取决于您。

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']