我想创建如下网址:
/api/foodfeeds/?keywords=BURGER,teste&mood=happy&location=2323,7767.323&price=2
urls.py
urlpatterns = [
path('admin/', admin.site.urls),
url(r'^api/foodfeed/(?P<keywords>[0-9.a-z, ]+)/(?P<mood>[0-9.a-z, ]+)/(?P<location>[0-9]+)/(?P<price>[0-9]+)/$', backend_views.FoodfeedList.as_view()),
]+ static(settings.STATIC_URL, document_root=settings.STATIC_ROOT)
views.py
class FoodfeedList(APIView):
# permission_classes = (permissions.IsAuthenticated,)
def get(self,request,keywords,mood,location,price):
print(request.GET['keywords'])
答案 0 :(得分:4)
正如@Umair所说,您将这些键作为 URL查询参数 传递,因此您不必在 {{1} }
就您而言,要获取通过URL传递的数据,请遵循以下代码段
URLPATTERNS
答案 1 :(得分:2)
new_df =
0 0
1 60
2 300
3 320
4 0
5 60
6 300
7 320
8 0
9 60
10 300
11 320
,keywords
,mood
等是location
,您不应在url中包括它们,而应通过query params
访问它们>
参考:http://www.django-rest-framework.org/api-guide/requests/#query_params