检查一个函数是否有一个装饰器 [英] Check if a function has a decorator
问题描述
我的问题是一般的,但我的应用程序是Django的login_required装饰器。
如果有办法检查是否有视图,我很好奇/ function有一个特定的装饰器(在这种情况下是login_required装饰器)
我在登录用户后重定向,如果页面,我想重定向到主页面他们当前正在登录_请求的装饰器。
构建您自己的 login_required
装饰器,并将其标记为装饰的功能 - 可能是标记它的最佳位置将在 func_dict
中。
django.contrib.auth.decorators导入login_required为django_l_r
#在这里您定义自己的装饰器,名为`login_required`
#它使用Django的内置`login_required`装饰器
def login_required(func):
decorated_func = django_l_r(func)
decorated_func.func_dict ['login_is_required'] = True
return decorated_func
@login_required#你的装饰器
def authenticatedd_view(request):
pass
def unauthenticated_view(request):
pass
现在你可以检查一下视图是否像这样装饰...
#假设`a_view`是查看函数
& GT;>> a_view.func_dict.get('login_is_required',False)
如果您对Python装饰器感到困惑,请参阅此SO问题/答案:如何使功能装饰器链?
My question is a general one, but specifically my application is the login_required decorator for Django.
I'm curious if there is a way to check if a view/function has a specific decorator (in this case the login_required decorator)
I am redirecting after logging a user out, and I want to redirect to the main page if the page they are currently on has the login_required decorator. My searches have yielded no results so far.
Build your own login_required
decorator and have it mark the function as decorated--probably the best place to mark it would be in the func_dict
.
from django.contrib.auth.decorators import login_required as django_l_r
# Here you're defining your own decorator called `login_required`
# it uses Django's built in `login_required` decorator
def login_required(func):
decorated_func = django_l_r(func)
decorated_func.func_dict['login_is_required'] = True
return decorated_func
@login_required # Your decorator
def authenticatedd_view(request):
pass
def unauthenticated_view(request):
pass
Now you can check to see if a view was decorated like this...
# Assume `a_view` is view function
>>> a_view.func_dict.get('login_is_required',False)
If you're confused about Python decorators see this SO question/answer: How to make a chain of function decorators?
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