Django REST框架+ Django REST Swagger + ImageField

Question

我创建了一个带有ImageField的简单模型,我想用django-rest-framework + django-rest-swagger创建一个api视图,该文档已记录并能够上传文件.

这是我得到的:

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()),
)

对我来说这应该很简单.但是,我无法使上传工作.我总是得到这个错误响应:

我从django-rest-framework中读到了这部分文档:

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.

但是,Header在Request Payload属性中从django-rest-swagger传递(来自chrome控制台).

如果需要更多信息,请告诉我.

我正在使用Django==1.8.8,djangorestframework==3.3.2而且django-rest-swagger==0.3.4.

Answer 1

我通过对代码进行一些更改来实现这一点.

首先,在models.py,更改ImageField名称file并使用上传文件夹的相对路径.当您将文件作为二进制流上传时,它可以在request.data文件key(request.data.get('file'))下的字典中使用,因此最干净的选项是将其映射到具有相同名称的模型字段.

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)

在serializer.py,将源字段重命名为file:

class MyModelSerializer(serializers.ModelSerializer):

    class Meta:
        model = MyModel
        fields = ('id', 'file', 'created_at')

在views.py中,不要调用super,而是调用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)

我使用Postman Chrome扩展程序来测试它.我已将图像作为二进制文件上传,我手动设置了两个标题:

Content-Disposition: attachment; filename=upload.jpg
Content-Type: */*

Answer 2

这是我想出的最终解决方案：

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)

我要做的就是将解析器从更改FileUploadParser为(FormParser, MultiPartParser)