Django REST 框架 + Django REST Swagger + ImageField

2022-01-22 00:00:00 python django django-rest-framework swagger django-swagger

问题描述

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

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.

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:

For me this should be pretty simple. However, I can't make the upload work. I always get this error response:

我已经从 django-rest 阅读了这部分文档-框架:

如果与 FileUploadParser 一起使用的视图使用文件名 URL 关键字参数调用，则该参数将用作文件名.如果在没有文件名 URL 关键字参数的情况下调用它，则客户端必须在 Content-Disposition HTTP 标头中设置文件名.例如 Content-Disposition:附件；文件名=upload.jpg.

但是，标题是由 django-rest-swagger 在 Request Payload 属性中传递的(来自 chrome 控制台).

However the Header is being passed by django-rest-swagger in the Request Payload property (from chrome console).

如果需要更多信息，请告诉我.

If any more info is necessary, please let me know.

我正在使用 Django==1.8.8、djangorestframework==3.3.2 和 django-rest-swagger==0.3.4.

I'm using Django==1.8.8, djangorestframework==3.3.2 and django-rest-swagger==0.3.4.

解决方案

这是我想出的最终解决方案:

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)

我所要做的就是将解析器从 FileUploadParser 更改为 (FormParser, MultiPartParser)

All I had to do was change the parsers from FileUploadParser to (FormParser, MultiPartParser)

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