我不能 post 数据 - Django
I cannot post data - Django
我目前正在构建 Django React 应用程序,我希望用户能够从前端创建模型(列表)的实例。
我试图构建后端,但是当我在 Postman 上尝试 url: http://localhost:8000/api/listings/create 时,我收到一个错误:“detail”:“Method”POST“不允许。”
此外,如果我改用“/listings/admin/create”,则不会出现此错误。
我的模型:
class Listing(models.Model):
artist = models.ForeignKey(Artist, on_delete=models.DO_NOTHING)
slug = models.CharField(max_length=200, unique=True)
title = models.CharField(max_length=150)
photo_main = models.ImageField(upload_to='photos/%Y/%m/%d/', default='photos/default.jpg')
description = models.TextField(blank=True)
def __str__(self):
return self.title
我的看法:
class CreateListingView(APIView):
parser_classes = [MultiPartParser, FormParser]
queryset = Listing.objects.all()
serializer_class = ListingSerializer
def post(self, request, format=None):
print(request.data)
serializer = ListingSerializer(data=request.data)
if serializer.is_valid():
serializer.save()
return Response(serializer.data,status=status.HTTP_200_OK)
else:
return Response(serializer.data,status=status.HTTP_400_BAD_REQUEST)
我的应用程序网址:
from django.urls import path
from .views import ListingsView, ListingView, SearchView,CreateListingView
app_name = 'listings'
urlpatterns = [
path('', ListingsView.as_view(),name='listings'),
path('search', SearchView.as_view(),name='listingssearc'),
path('<slug>', ListingView.as_view(),name='listingsdetails'),
path('create', CreateListingView.as_view(),name='listingscreate')
]
我的项目urls :
urlpatterns = [
path('api-auth/', include('rest_framework.urls')),
path('api/token/', TokenObtainPairView.as_view(), name='token_obtain_pair'),
path('api/token/refresh/', TokenRefreshView.as_view(), name='token_refresh'),
path('api/accounts/', include('accounts.urls')),
path('api/artists/', include('artists.urls')),
path('api/listings/', include('listings.urls', namespace='listings')),
path('api/contact/', include('contact.urls')),
path('admin/', admin.site.urls),
] + static(settings.MEDIA_URL, document_root=settings.MEDIA_ROOT)
urlpatterns += [re_path(r'^.*', TemplateView.as_view(template_name='index.html'))]
那我做错了什么?我不明白为什么会出现 405 错误。
感谢阅读!
应该是这样的:
class CreateListingView(APIView):
def post(self, request, format=None):
print(request.data)
serializer = ListingSerializer(data=request.data)
if serializer.is_valid():
serializer.save()
return Response(serializer.data,status=status.HTTP_200_OK)
else:
return Response(serializer.data,status=status.HTTP_400_BAD_REQUEST)
有关参考,请参阅此 drf APIView
你可以尝试改变
class CreateListingView(APIView):
到
class CreateListingView(generics.ListCreateAPIView)
url-pattern一行接一行地尝试。
在您的 urlpatterns 中,您在之前的行中捕捉到对“创建”的调用作为 slug:
urlpatterns = [
path('', ListingsView.as_view(),name='listings'),
path('search', SearchView.as_view(),name='listingssearc'),
path('<slug>', ListingView.as_view(),name='listingsdetails'),
path('create', CreateListingView.as_view(),name='listingscreate')
]
所以它将进入列表详细信息,这显然不允许 post 请求。
改为:
urlpatterns = [
path('', ListingsView.as_view(),name='listings'),
path('search', SearchView.as_view(),name='listingssearc'),
path('create', CreateListingView.as_view(),name='listingscreate'),
path('<slug>', ListingView.as_view(),name='listingsdetails')
]
如果它有一个不是搜索或创建的 slug,它只会被最后一行捕获。
我目前正在构建 Django React 应用程序,我希望用户能够从前端创建模型(列表)的实例。 我试图构建后端,但是当我在 Postman 上尝试 url: http://localhost:8000/api/listings/create 时,我收到一个错误:“detail”:“Method”POST“不允许。”
此外,如果我改用“/listings/admin/create”,则不会出现此错误。
我的模型:
class Listing(models.Model):
artist = models.ForeignKey(Artist, on_delete=models.DO_NOTHING)
slug = models.CharField(max_length=200, unique=True)
title = models.CharField(max_length=150)
photo_main = models.ImageField(upload_to='photos/%Y/%m/%d/', default='photos/default.jpg')
description = models.TextField(blank=True)
def __str__(self):
return self.title
我的看法:
class CreateListingView(APIView):
parser_classes = [MultiPartParser, FormParser]
queryset = Listing.objects.all()
serializer_class = ListingSerializer
def post(self, request, format=None):
print(request.data)
serializer = ListingSerializer(data=request.data)
if serializer.is_valid():
serializer.save()
return Response(serializer.data,status=status.HTTP_200_OK)
else:
return Response(serializer.data,status=status.HTTP_400_BAD_REQUEST)
我的应用程序网址:
from django.urls import path
from .views import ListingsView, ListingView, SearchView,CreateListingView
app_name = 'listings'
urlpatterns = [
path('', ListingsView.as_view(),name='listings'),
path('search', SearchView.as_view(),name='listingssearc'),
path('<slug>', ListingView.as_view(),name='listingsdetails'),
path('create', CreateListingView.as_view(),name='listingscreate')
]
我的项目urls :
urlpatterns = [
path('api-auth/', include('rest_framework.urls')),
path('api/token/', TokenObtainPairView.as_view(), name='token_obtain_pair'),
path('api/token/refresh/', TokenRefreshView.as_view(), name='token_refresh'),
path('api/accounts/', include('accounts.urls')),
path('api/artists/', include('artists.urls')),
path('api/listings/', include('listings.urls', namespace='listings')),
path('api/contact/', include('contact.urls')),
path('admin/', admin.site.urls),
] + static(settings.MEDIA_URL, document_root=settings.MEDIA_ROOT)
urlpatterns += [re_path(r'^.*', TemplateView.as_view(template_name='index.html'))]
那我做错了什么?我不明白为什么会出现 405 错误。
感谢阅读!
应该是这样的:
class CreateListingView(APIView):
def post(self, request, format=None):
print(request.data)
serializer = ListingSerializer(data=request.data)
if serializer.is_valid():
serializer.save()
return Response(serializer.data,status=status.HTTP_200_OK)
else:
return Response(serializer.data,status=status.HTTP_400_BAD_REQUEST)
有关参考,请参阅此 drf APIView
你可以尝试改变
class CreateListingView(APIView):
到
class CreateListingView(generics.ListCreateAPIView)
url-pattern一行接一行地尝试。
在您的 urlpatterns 中,您在之前的行中捕捉到对“创建”的调用作为 slug:
urlpatterns = [
path('', ListingsView.as_view(),name='listings'),
path('search', SearchView.as_view(),name='listingssearc'),
path('<slug>', ListingView.as_view(),name='listingsdetails'),
path('create', CreateListingView.as_view(),name='listingscreate')
]
所以它将进入列表详细信息,这显然不允许 post 请求。
改为:
urlpatterns = [
path('', ListingsView.as_view(),name='listings'),
path('search', SearchView.as_view(),name='listingssearc'),
path('create', CreateListingView.as_view(),name='listingscreate'),
path('<slug>', ListingView.as_view(),name='listingsdetails')
]
如果它有一个不是搜索或创建的 slug,它只会被最后一行捕获。