Python models.Permission方法代碼示例

# 需要導入模塊: from django.contrib.auth import models [as 別名]
# 或者: from django.contrib.auth.models import Permission [as 別名]
def settings(self):
        if self.cached_settings:
            return self.cached_settings
        else:
            from django.conf import settings
            allowed_configurations = {
                'CONTENT_TYPE_CLASS': ContentType,
                'USER_CLASS': settings.AUTH_USER_MODEL,
                'PERMISSION_CLASS': Permission,
                'GROUP_CLASS': Group,
                'INJECT_MODEL_ADMIN': False
            }
            river_settings = {}
            for key, default in allowed_configurations.items():
                river_settings[key] = getattr(settings, self.get_with_prefix(key), default)

            river_settings['IS_MSSQL'] = connection.vendor == 'microsoft'
            self.cached_settings = river_settings

            return self.cached_settings 

# 需要導入模塊: from django.contrib.auth import models [as 別名]
# 或者: from django.contrib.auth.models import Permission [as 別名]
def setUp(self):
        super(ReadTestCase, self).setUp()

        self.group = Group.objects.create(name='test_group')
        self.group__permissions = list(Permission.objects.all()[:3])
        self.group.permissions.add(*self.group__permissions)
        self.user = User.objects.create_user('user')
        self.user__permissions = list(Permission.objects.all()[3:6])
        self.user.groups.add(self.group)
        self.user.user_permissions.add(*self.user__permissions)
        self.admin = User.objects.create_superuser('admin', 'admin@test.me',
                                                   'password')
        self.t1__permission = (Permission.objects.order_by('?')
                               .select_related('content_type')[0])
        self.t1 = Test.objects.create(
            name='test1', owner=self.user,
            date='1789-07-14', datetime='1789-07-14T16:43:27',
            permission=self.t1__permission)
        self.t2 = Test.objects.create(
            name='test2', owner=self.admin, public=True,
            date='1944-06-06', datetime='1944-06-06T06:35:00') 

# 需要導入模塊: from django.contrib.auth import models [as 別名]
# 或者: from django.contrib.auth.models import Permission [as 別名]
def test_filtered_relation(self):
        from django.db.models import FilteredRelation

        qs = TestChild.objects.annotate(
            filtered_permissions=FilteredRelation(
                'permissions', condition=Q(permissions__pk__gt=1)))
        self.assert_tables(qs, TestChild)
        self.assert_query_cached(qs)

        values_qs = qs.values('filtered_permissions')
        self.assert_tables(
            values_qs, TestChild, TestChild.permissions.through, Permission)
        self.assert_query_cached(values_qs)

        filtered_qs = qs.filter(filtered_permissions__pk__gt=2)
        self.assert_tables(
            values_qs, TestChild, TestChild.permissions.through, Permission)
        self.assert_query_cached(filtered_qs) 

# 需要導入模塊: from django.contrib.auth import models [as 別名]
# 或者: from django.contrib.auth.models import Permission [as 別名]
def test_difference(self):
        qs = Test.objects.filter(pk__lt=5)
        sub_qs = Test.objects.filter(permission__name__contains='a')
        if self.is_sqlite:
            qs = qs.order_by()
            sub_qs = sub_qs.order_by()
        qs = qs.difference(sub_qs)
        self.assert_tables(qs, Test, Permission)
        self.assert_query_cached(qs)

        qs = Test.objects.all()
        sub_qs = Permission.objects.all()
        if self.is_sqlite:
            qs = qs.order_by()
            sub_qs = sub_qs.order_by()
        qs = qs.difference(sub_qs)
        self.assert_tables(qs, Test, Permission)
        with self.assertRaises((ProgrammingError, OperationalError)):
            self.assert_query_cached(qs) 

# 需要導入模塊: from django.contrib.auth import models [as 別名]
# 或者: from django.contrib.auth.models import Permission [as 別名]
def _get_users_with_any_permission_codenames_filter(self, permission_codenames):
        """
        Given a list of permission codenames, return a filter expression which
        will find all users which have any of those permissions - either
        through group permissions, user permissions, or implicitly through
        being a superuser.
        """
        permissions = Permission.objects.filter(
            content_type=self._content_type,
            codename__in=permission_codenames
        )
        return (
            Q(is_superuser=True)
            | Q(user_permissions__in=permissions)
            | Q(groups__permissions__in=permissions)
        ) & Q(is_active=True) 

# 需要導入模塊: from django.contrib.auth import models [as 別名]
# 或者: from django.contrib.auth.models import Permission [as 別名]
def setUp(self):
        self.TestGroup = get_default_group_ctype().model_class()
        self.user = TestUser.objects.create(username='test1', email='test@test.com')
        self.user2 = TestUser.objects.create(username='test2', email='test2@test.com')
        self.user3 = TestUser.objects.create(username='test3', email='test3@test.com')
        self.responsible_user = TestUser.objects.create_user(username='responsible')
        self.permission = Permission.objects.create(
            codename='test', content_type=get_user_ctype()
        )
        self.permission2 = Permission.objects.create(
            codename='test2', content_type=get_user_ctype()
        )
        self.permission_key = get_user_ctype().app_label + '.test'
        self.permission2_key = get_user_ctype().app_label + '.test2'
        self.group = self.TestGroup.objects.create(
            name='test_group'
        )
        self.group2 = self.TestGroup.objects.create(
            name='test_group2'
        )
        self.group2.restrict()
        self.group2.save()
        self.HistoryOwnerToPermission = HistoryOwnerToPermission
        self.HistoryGenericUserToGroup = HistoryGenericUserToGroup 

# 需要導入模塊: from django.contrib.auth import models [as 別名]
# 或者: from django.contrib.auth.models import Permission [as 別名]
def __unicode__(self):
        if self.object_id is None:
            ctype = None
        else:
            ctype = self.content_type
        result = "{app}.{model}.{pk} ".format(
            app=self.owner_content_type.app_label,
            model=self.owner_content_type.model,
            pk=self.owner_object_id,
        )
        if self.object_id is not None:  # real object not global permission
            result += "- {app}.{model}.{pk}. ".format(
                app=ctype.app_label if ctype else '',
                model=ctype.model if ctype else '',
                pk=self.object_id or '',
            )
        if self.roles:
            result += "Roles {roles}. ".format(roles=self.roles)
        result += "Permission {perm}".format(perm=self.permission.codename)
        return result 

# 需要導入模塊: from django.contrib.auth import models [as 別名]
# 或者: from django.contrib.auth.models import Permission [as 別名]
def _filter_anonymous_perms(self, unfiltered_set):
        """
        Restrict a set of tuples in the format (user_id, permission_id) to
        only those permissions that apply to the content_type of this object
        and are listed in settings.ALLOWED_ANONYMOUS_PERMISSIONS.
        """
        content_type = ContentType.objects.get_for_model(self)
        # Translate settings.ALLOWED_ANONYMOUS_PERMISSIONS to primary keys
        codenames = set()
        for perm in settings.ALLOWED_ANONYMOUS_PERMISSIONS:
            app_label, codename = perm_parse(perm)
            if app_label == content_type.app_label:
                codenames.add(codename)
        allowed_permission_ids = Permission.objects.filter(
            content_type_id=content_type.pk, codename__in=codenames
        ).values_list('pk', flat=True)
        filtered_set = copy.copy(unfiltered_set)
        for user_id, permission_id in unfiltered_set:
            if user_id == settings.ANONYMOUS_USER_ID:
                if permission_id not in allowed_permission_ids:
                    filtered_set.remove((user_id, permission_id))
        return filtered_set 

# 需要導入模塊: from django.contrib.auth import models [as 別名]
# 或者: from django.contrib.auth.models import Permission [as 別名]
def get_users_with_perms(self, attach_perms=False):
        """ Return a QuerySet of all users with any effective grant permission
        on this object. If attach_perms=True, then return a dict with
        users as the keys and lists of their permissions as the values. """
        user_perm_ids = self._get_effective_perms()
        if attach_perms:
            user_perm_dict = {}
            for user_id, perm_id in user_perm_ids:
                perm_list = user_perm_dict.get(user_id, [])
                perm_list.append(Permission.objects.get(pk=perm_id).codename)
                user_perm_dict[user_id] = sorted(perm_list)
            # Resolve user ids into actual user objects
            user_perm_dict = {User.objects.get(pk=key): value for key, value
                              in user_perm_dict.items()}
            return user_perm_dict
        else:
            # Use a set to avoid duplicate users
            user_ids = {x[0] for x in user_perm_ids}
            return User.objects.filter(pk__in=user_ids) 

# 需要導入模塊: from django.contrib.auth import models [as 別名]
# 或者: from django.contrib.auth.models import Permission [as 別名]
def reset_group_permissions(group_name, app_permissions=(), cms_permissions=(), sites=()):
    """
    Initialize a Django auth group assigning a set of Django app permissions,
    and global page permissions for django CMS.

    :param group_name: Name of Django auth group to create or reset
    :param app_permissions: list of Django auth Permission objects
    :param cms_permissions: dictionary of django CMS permissions
    :param sites: optional list of Django sites objects (None: all sites)
    :return: the created or updated group model instance
    """
    group, created = Group.objects.get_or_create(name=group_name)
    group.permissions = permission_list(app_permissions)
    group.save()

    perms, created = GlobalPagePermission.objects.get_or_create(group=group)
    for attrib, value in dict(cms_permissions).items():
        setattr(perms, attrib, value)
    perms.sites = sites
    perms.save()

    return group 

# 需要導入模塊: from django.contrib.auth import models [as 別名]
# 或者: from django.contrib.auth.models import Permission [as 別名]
def remove_perm_from_user(user, perm):
    """Remove a permission from an user"""

    if isinstance(perm, str):
        try:
            app_label, codename = perm.split('.')
        except ValueError:
            raise ValueError('%s is not valid. Should be in format app_label.perm_codename')
        else:
            if not app_label or not codename:
                raise ValueError('Invalid app_label or codename')
            get_permission = Permission.objects.get
            user.user_permissions.remove(
                get_permission(content_type__app_label=app_label, codename=codename))
    elif isinstance(perm, Permission):
        user.user_permissions.remove(perm)
    else:
        raise TypeError('perm should be an instance of either str or Permission') 

# 需要導入模塊: from django.contrib.auth import models [as 別名]
# 或者: from django.contrib.auth.models import Permission [as 別名]
def test_default_permissions(self):
        permission_content_type = ContentType.objects.get_by_natural_key('auth', 'permission')
        Permission._meta.permissions = [
            ('my_custom_permission', 'Some permission'),
        ]
        create_permissions(self.app_config, verbosity=0)

        # view/add/change/delete permission by default + custom permission
        self.assertEqual(Permission.objects.filter(
            content_type=permission_content_type,
        ).count(), 5)

        Permission.objects.filter(content_type=permission_content_type).delete()
        Permission._meta.default_permissions = []
        create_permissions(self.app_config, verbosity=0)

        # custom permission only since default permissions is empty
        self.assertEqual(Permission.objects.filter(
            content_type=permission_content_type,
        ).count(), 1) 

# 需要導入模塊: from django.contrib.auth import models [as 別名]
# 或者: from django.contrib.auth.models import Permission [as 別名]
def get_editable_permissions(self):
        """
        Return a queryset of Permission objects that can be assigned

        The view has an attribute called `editable_permissions` but that
        attribute only lists app names and permission codenames. We need to
        turn that tuple of tuples into a queryset of permissions.
        """
        # Dynamic generation of OR queries is based on code found at
        # https://bradmontgomery.net/blog/adding-q-objects-in-django/
        permission_filter = Q()
        for permission in self.editable_permissions:
            permission_filter.add(
                Q(content_type__app_label=permission[0],
                  codename=permission[1]), Q.OR)

        return Permission.objects.filter(
            permission_filter) 

# 需要導入模塊: from django.contrib.auth import models [as 別名]
# 或者: from django.contrib.auth.models import Permission [as 別名]
def get_permissions_queryset(self):
        """
        Get the queryset to pre-fill the optional individual permission list

        There are a significant number of possible permissions within every
        django app. Most of these permissions are not necessary for the
        operation of the app, so we never display more than the ones required.

        However, due to the way that UpdateViews and Many-to-Many forms are
        saved when a form with a many-to-many field is submitted all existing
        relationships are cleared and replaced with those submitted.

        If, by some chance, a user has an individual permission that is not
        listed above, the easiest way to handle this is to simply append the
        permission to the form in order to prevent data corruption.
        """
        editable_permissions_queryset = self.get_editable_permissions()
        existing_permissions_queryset = self.object.user_permissions.all()

        return Permission.objects.filter(
            Q(pk__in=editable_permissions_queryset.values('pk')) |
            Q(pk__in=existing_permissions_queryset.values('pk'))
            ).order_by('content_type__app_label').select_related('content_type') 

# 需要導入模塊: from django.contrib.auth import models [as 別名]
# 或者: from django.contrib.auth.models import Permission [as 別名]
def setUp(self):
        """Handy things."""
        self.request_factory = RequestFactory()

        # Add 2 permissions to the test, one valid and visible, one hidden
        demo_content_type = ContentType.objects.create(
            app_label='demo-app-label', model='DemoModel')

        self.valid_permission = mommy.make(
            Permission,
            codename='viewable-permission',
            name='Viewable Permission',
            content_type=demo_content_type)
        self.hidden_permission = mommy.make(
            Permission,
            codename='hidden-permission',
            name='Hidden Permission',
            content_type=demo_content_type)

        # Create a view class that contains those permissions
        self.view_class = views.UpdateUserPermissionView
        self.view_class.editable_permissions = (
            ('demo-app-label', 'viewable-permission'),
        )