格式化权限QuerySet，使其仅显示名称列

Question

我正在尝试通过查询列出特定于模型的权限。目标是仅显示QuerySet的“名称”部分。我只能通过以下标签获取输出：{{ permissions }}

<QuerySet [<Permission: cms | Account meta | Can add Account meta>, <Permission: cms | Account meta | Can change Account meta>, <Permission: cms | Account meta | Can delete Account meta>, <Permission: cms | Account meta | Allows the user to log in as others>, <Permission: cms | Account meta | Allows a user to deactivate/delete other users>, <Permission: cms | Account meta | Allows a user to edit site and role details>, <Permission: cms | Account meta | Can export user details>, <Permission: cms | Account meta | Can view Account meta>]>

models.py

class Account(models.Model):
    user = models.OneToOneField(User, on_delete=models.CASCADE)
    site = models.CharField(max_length=50, choices=(('all', 'all'), ('danielk', 'danielk'), ('flis', 'flis'), ('vmusic', 'vmusic')), blank=True)
    site_role = models.CharField(max_length=50, choices=(('Administrator', 'Administrator'), ('Moderator', 'Moderator'), ('Editor', 'Editor')))
    phone_number = models.CharField(max_length=8)
    birth_date = models.DateField()
    street_adress = models.CharField(max_length=255)
    note = models.TextField(blank=True);
    zip_code = models.CharField(max_length=4)
    city = models.CharField(max_length=255)

    def formatedPhone(self, country=None):
        return phonenumbers.parse(Account.phone_number, "NO")

    """
    def __str__(self):
        return "%s %s" % (self.User.first_name, self.user.last_name)
    """

    def get_absolute_url(self):
        return reverse('account-detail', kwargs={'pk': self.pk})

    class Meta:
        verbose_name = 'Account meta'
        verbose_name_plural = 'Accounts meta'
        permissions = (
            ("has_user_hijack_permission", "Allows the user to log in as others"),
            ("has_user_management", "Allows a user to deactivate/delete other users"),
            ("has_user_site_edit", "Allows a user to edit site and role details"),
            ("has_user_takeout", "Can export user details"),
        )

views.py

class cms_users_user_permissions(generic.DetailView):
    model = Account
    template_name = 'cms/users/cms-users-user-permissions.html'

    def get_context_data(self, **kwargs):
        context = super().get_context_data(**kwargs)
        context["permissions"] = Permission.objects.filter(content_type=ContentType.objects.get_for_model(Account))
        #context['permissions'] = Permission.objects.filter(content_type_id=7)
        return context

Answer 1

您无需遍历{{ permissions }}，而是遍历查询集，并每次打印权限名称，例如：

{% for permission in permissions %}
    {{ permission.name }}
{% endfor %}