我正在使用django user_passes_test decorator检查用户权限。
@user_passes_test(lambda u: has_add_permission(u, "project"))
def create_project(request):
......
我正在调用一个回调函数,它具有带两个参数user和一个字符串的\添加\权限。我想将请求对象与它一起传递,这是可能的吗?另外,有谁能告诉我我们如何能够直接访问decorator内部的用户对象吗?
最佳答案
不,您不能将请求传递给user_passes_test
。要了解其工作原理和原因,只需前往:
def user_passes_test(test_func, login_url=None, redirect_field_name=REDIRECT_FIELD_NAME):
"""
Decorator for views that checks that the user passes the given test,
redirecting to the log-in page if necessary. The test should be a callable
that takes the user object and returns True if the user passes.
"""
def decorator(view_func):
@wraps(view_func, assigned=available_attrs(view_func))
def _wrapped_view(request, *args, **kwargs):
if test_func(request.user):
return view_func(request, *args, **kwargs)
path = request.build_absolute_uri()
# If the login url is the same scheme and net location then just
# use the path as the "next" url.
login_scheme, login_netloc = urlparse.urlparse(login_url or
settings.LOGIN_URL)[:2]
current_scheme, current_netloc = urlparse.urlparse(path)[:2]
if ((not login_scheme or login_scheme == current_scheme) and
(not login_netloc or login_netloc == current_netloc)):
path = request.get_full_path()
from django.contrib.auth.views import redirect_to_login
return redirect_to_login(path, login_url, redirect_field_name)
return _wrapped_view
return decorator
这是装饰器后面的代码。如您所见,传递给decorator的测试函数(在您的例子中,
user_passes_test
)只传递一个参数,lambda u: has_add_permission(u, "project")
。现在,当然可以编写自己的decorator(甚至直接复制此代码并对其进行修改)来传递request.user
本身,但不能使用默认的request
实现。关于python - 如何在user_passes_test装饰器可调用函数中传递Django请求对象,我们在Stack Overflow上找到一个类似的问题:https://stackoverflow.com/questions/11872560/