如何在传递给manage.py的参数函数中更改urls.py中的urlpatterns路径？

Question

我要更改urls.py：

urlpatterns = [path（'rmt / {}'。format（url），view）]

从传递给manage.py的参数

class InitializeService():
    def __init__(self):
        os.environ.setdefault('DJANGO_SETTINGS_MODULE', 'RMT_Django.settings')
        try:
            from django.core.management import execute_from_command_line
        except ImportError as exc:
            raise ImportError(
                "Couldn't import Django. Are you sure it's installed and "
                "available on your PYTHONPATH environment variable? Did you "
                "forget to activate a virtual environment?"
            ) from exc

            flag_env = sys.argv[6]
            if flag_env == "DEV":
                url = "/rmt/push_dev"
            elif flag_env == "TEST":
                url = "/rmt/push_test"
            elif flag_env == "PROD":
                url = "/rmt/push"

        execute_from_command_line(sys.argv[0:3])

InitializeService()

如何将url传递给urlpattern？

谢谢

Answer 1

我认为这不是正确的方法。

方法1：在views.py中进行

# urls.py

urlpatterns = [
    path('rmt/<str:service_url>/'.format(url), handeller_view)
]

# views.py

flag_env = sys.argv[6]
service_mapping = {'DEV': 'push_dev', 'TEST': 'push_test', 'PROD': 'push'}

def handeller_view(request, service_url):
    if service_url == service_mapping[flag_env]:
        # handel your request accordingly

    raise Http404()

方法2：在urls.py中进行

或者，您可以根据args来匹配URL。

# urls.py

flag_env = sys.argv[6]
if flag_env == "DEV":
   url = "/rmt/push_dev"
elif flag_env == "TEST":
   url = "/rmt/push_test"
elif flag_env == "PROD":
   url = "/rmt/push"


urlpatterns = [
   path(url, view),
]

或者您可以根据您的标志创建一个URL模式，如果根据您的标志有多个URL，这将很有用。

# urls.py

flag_env = sys.argv[6]
if flag_env == "DEV":
   urlpatterns = [
        path("/rmt/push_dev", view),
    ]
elif flag_env == "TEST":
   urlpatterns = [
        path("/rmt/push_test", view),
    ]

elif flag_env == "PROD":
   urlpatterns = [
        path("/rmt/push", view),
    ]