[英]Django URL, pass parameters in URL
我想創建如下網址:
/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'])
正如@Umair所說,您要將這些鍵作為URL查詢參數進行傳遞,因此不必在URLPATTERNS
提及它
對於您的情況,要獲取通過URL傳遞的數據,請遵循以下代碼段
#urls.py
urlpatterns = [
path('admin/', admin.site.urls),
url(r'^api/foodfeed/', 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): print(request.GET) # print all url params print(request.GET['keywords']) print(request.GET['mood'])
# etc
這些keywords
, mood
, location
等是query params
您不應在url中包括這些query params
,而應通過request.query_params
訪問它們
參考: http : //www.django-rest-framework.org/api-guide/requests/#query_params
聲明:本站的技術帖子網頁,遵循CC BY-SA 4.0協議,如果您需要轉載,請注明本站網址或者原文地址。任何問題請咨詢:yoyou2525@163.com.