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目前我传递参数“ID”在URL中调用API和我打电话的ID的API,但是我想通过parameter.here调用API是views.py`Django的RestFramework由现场
class post_list(APIView):
def get(self,request,format=None):
post_resource=PostResource()
dataset=Dataset()
post = Post.objects.all()
serializer = PostSerializer(post, many=True)
a=[]
for row in post:
a.append(row)
return Response(serializer.data)
def post(self,request,format =None):
serializer = PostSerializer(data=request.data)
if serializer.is_valid():
serializer.save()
return Response(serializer.data, status=201)
return Response(serializer.errors, status=400)
class post_detail(APIView):
def get_object(self, pk):
try:
return Post.objects.get(pk=pk)
except Post.DoesNotExist:
raise Http404
def get(self, request, pk, format=None):
post = self.get_object(pk)
serializer = PostSerializer(post)
return Response(serializer.data)
def put(self, request, pk, format=None):
snippet = self.get_object(pk)
serializer = PostSerializer(snippet, data=request.data)
if serializer.is_valid():
serializer.save()
return Response(serializer.data)
return Response(serializer.errors, status=status.HTTP_400_BAD_REQUEST)
def delete(self, request, pk, format=None):
post = self.get_object(pk)
post.delete()
return Response(status=status.HTTP_204_NO_CONTENT)
当我通过ID在它的只显示该特定ID结果网址不过,我需要视图如果我通过其他领域inste广告的id它显示的结果基于that.so我的问题是我应该改变views.py和Urls.py。这是urls.py file`
from django.conf.urls import url
from api import views
from rest_framework.urlpatterns import format_suffix_patterns
urlpatterns = [
url(r'^post/$', views.post_list.as_view()),
url(r'^post/(?P<pk>[0-9]+)/$', views.post_detail.as_view()),
]
urlpatterns=format_suffix_patterns(urlpatterns)
这是我serializer.py file`
from rest_framework import serializers
from .models import Post
class PostSerializer(serializers.ModelSerializer):
class Meta:
model = Post
#fields=('ProductName','Score')
fields ='__all__'
`
这取决于你想传递什么而不是'id'。它是什么类型?这是一个字符串吗?或“uuid”? –
我想通过字符串“PName”。 – Nishit
最有可能你想要查询参数,以便您可以通过其他字段进行过滤。 –