Question

我正在关注Django-Restful教程here，我编写了以下代码

在views.py中

：

from snippets.models import Snippet                                              
from snippets.serializers import SnippetSerializer                               

from rest_framework import status                                                
from rest_framework.decorators import api_view                                   
from rest_framework.response import Response                                     

@api_view(['GET','POST'])                                                        
def snippet_list(request, format=None):                                          
    """                                                                          
    List all snippets, or create a new snippet.                                  
    """                                                                          
    if request.method == 'GET':                                                  
        snippets = Snippet.objects.all()                                         
        serializer = SnippetSerializer(snippets, many=True)                      
        return Response(serializer.data)                                         
    elif request.method == 'POST':                                               
        serializer = SnippetSerializer(data=request.data)                        
        if serializer.is_valid():                                                
            serializer.save()                                                    
            return Response(serializer.data,status=status.HTTP_201_CREATED)      
        return Response(serializer.errors, status=status.HTTP_400_BAD_REQUEST)

在urls.py中

：

from django.conf.urls import url
from snippets import views
from rest_framework.urlpatterns import format_suffix_patterns

urlpatterns = [
    url(r'^snippets/$',views.snippet_list),
    url(r'^snippets/(?P<pk>[0-9]+)/$',views.snippet_detail)
]

urlpatterns = format_suffix_patterns(urlpatterns)

当我运行服务器时，我可以获得正常的http://127.0.0.1:8000/snippets/来返回标准的JSON响应，但是当我尝试访问时

http http://127.0.0.1:8000/snippets.json  # JSON suffix
http http://127.0.0.1:8000/snippets.api   # Browsable API suffix

我收到一个产生此追溯的错误：

Environment:


Request Method: GET
Request URL: http://127.0.0.1:8000/snippets.json

Django Version: 1.9.7
Python Version: 2.7.6
Installed Applications:
['django.contrib.admin',
 'django.contrib.auth',
 'django.contrib.contenttypes',
 'django.contrib.sessions',
 'django.contrib.messages',
 'django.contrib.staticfiles',
 'rest_framework',
 'snippets',
 'quickstart']
Installed 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']



Traceback:

File "/home/myname/workspace/restTutorial/env/local/lib/python2.7/site-packages/django/core/handlers/base.py" in get_response
  149.                     response = self.process_exception_by_middleware(e, request)

File "/home/myname/workspace/restTutorial/env/local/lib/python2.7/site-packages/django/core/handlers/base.py" in get_response
  147.                     response = wrapped_callback(request, *callback_args, **callback_kwargs)

File "/home/myname/workspace/restTutorial/env/local/lib/python2.7/site-packages/django/views/decorators/csrf.py" in wrapped_view
  58.         return view_func(*args, **kwargs)

Exception Type: TypeError at /snippets.json
Exception Value: snippet_list() got an unexpected keyword argument 'format'

为什么我收到此错误？

修改：指向settings.py

的链接

Answer 1

问题是我在views.py文件中有重复的函数，其中不同的装饰器靠近底部。删除它们解决了这个问题。

Answer 2

试试这个

http http://127.0.0.1:8000/snippets.json  # JSON suffix http
http://127.0.0.1:8000/snippets.api   # Browsable API suffix

而不是

http http://127.0.0.1:8000/snippets/?format=json  
http http://127.0.0.1:8000/snippets/?format=api

Answer 3

我遇到一个类似的问题，我的错误是我在声明URL而不是Django Rest Framework的反向模块时使用的是Django的反向模块：

请勿使用此：

from django.urls import reverse

改为使用此：

from rest_framework.reverse import reverse

@api_view(['GET'])
def api_root(request, format=None):
    return Response({
       #'users': reverse('users:user-list', request=request, format=format),
       'transactions': reverse('transactions:transaction-list', request=request, format=format),
})

为什么这个Django API调用会出现意外的关键字参数错误？

3 个答案: