Fixed #2629 -- Added a permission_required decorator to
django.contrib.auth.decorator. Thanks, dummy@habmalnefrage.de. git-svn-id: http://code.djangoproject.com/svn/django/trunk@3779 bcc190cf-cafb-0310-a4f2-bffc1f526a37
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@ -26,3 +26,11 @@ login_required.__doc__ = (
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to the log-in page if necessary.
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"""
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)
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def permission_required(perm, login_url=LOGIN_URL):
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"""
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Decorator for views that checks if a user has a particular permission
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enabled, redirectiing to the log-in page if necessary.
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"""
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return user_passes_test(lambda u: u.has_perm(perm), login_url=login_url)
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@ -456,6 +456,10 @@ As a shortcut, you can use the convenient ``user_passes_test`` decorator::
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# ...
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my_view = user_passes_test(lambda u: u.has_perm('polls.can_vote'))(my_view)
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We are using this particular test as a relatively simple example, however be
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aware that if you just want to test if a permission is available to a user,
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you can use the ``permission_required()`` decorator described below.
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Here's the same thing, using Python 2.4's decorator syntax::
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from django.contrib.auth.decorators import user_passes_test
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@ -488,6 +492,24 @@ Example in Python 2.4 syntax::
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def my_view(request):
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# ...
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The permission_required decorator
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~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
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Since checking whether a user has a particular permission available to them is a
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relatively common operation, Django provides a shortcut for that particular
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case: the ``permission_required()`` decorator. Using this decorator, the
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earlier example can be written as::
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from django.contrib.auth.decorators import permission_required
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def my_view(request):
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# ...
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my_view = permission_required('polls.can_vote')(my_view)
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Note that ``permission_required()`` also takes an optional ``login_url``
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parameter.
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Limiting access to generic views
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--------------------------------
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