Re: Read Only by user in Django Admin

Hi,

The following PR is ready to merge and implements this: https://github.com/django/django/pull/6734

I will squash the commits  as soon as I'm back in the office (next week) and it will probably be merged soon.

Bests

Olivier

On 25 Oct 2016 19:03, "Luis Zárate" <luisza14@gmail.com> wrote:

you are looking something like
https://code.djangoproject.com/ticket/7150
https://code.djangoproject.com/attachment/ticket/7150/admin_view-permission-1.0.patch
https://code.djangoproject.com/attachment/ticket/7150/newforms-admin-view-permission-r7737.patch
https://code.djangoproject.com/ticket/8936
https://code.djangoproject.com/ticket/22452
https://code.djangoproject.com/ticket/18814
https://code.djangoproject.com/ticket/17295
https://code.djangoproject.com/ticket/3984
https://code.djangoproject.com/ticket/2058
https://code.djangoproject.com/ticket/820

 

I don't know why Django doesn't support readonly view, various proposal was summit and always was reject. 

Django Admin needs Readonly View functionality for a complete implementation of CRUD.

There is a lot of scenarios where you need readonly view for your staff.  And yes you can implement a readonly  in django admin without modify django core admin with some tricks.

I

2016-10-24 7:26 GMT-06:00 Derek <gamesbook@gmail.com>:

Would it be possible to add these extra admin methods into a parent class; and then have all your individual model admins inherit from it?  Ditto for the models.py

On Sunday, 8 February 2015 21:15:42 UTC+2, Hangloser Firestarter wrote:

Solved.

In __init__.py
...
from django.db.models.signals import post_syncdb
from django.contrib.contenttypes.models import ContentType
from django.contrib.auth.models import Permission

def add_view_permissions(sender, **kwargs):
    """
    This syncdb hooks takes care of adding a view permission too all our
    content types.
    """
    # for each of our content types
    for content_type in ContentType.objects.all():
        # build our permission slug
        codename = "view_%s" % content_type.model

        # if it doesn't exist..
        if not Permission.objects.filter(content_type=content_type, codename=codename):
            # add it
            Permission.objects.create(content_type=content_type,
                                      codename=codename,
                                      name="Can view %s" % content_type.name)
            print("Added view permission for %s" % content_type.name)

# check for all our view permissions after a syncdb
post_syncdb.connect(add_view_permissions)
...

In admin.py
...
class MyAdmin(admin.ModelAdmin):
    def has_change_permission(self, request, obj=None):
        ct = ContentType.objects.get_for_model(self.model)
        salida = False
        if request.user.is_superuser:
            salida = True
        else:
            if request.user.has_perm('%s.view_%s' % (ct.app_label, ct.model)):
                salida = True
            else:
                if request.user.has_perm('%s.change_%s' % (ct.app_label, ct.model)):
                    salida = True
                else:
                    salida = False
        return salida

    def get_readonly_fields(self, request, obj=None):
        ct = ContentType.objects.get_for_model(self.model)
        if not request.user.is_superuser and request.user.has_perm('%s.view_%s' % (ct.app_label, ct.model)):
            return [el.name for el in self.model._meta.fields]
        return self.readonly_fields
...

in models.py
...
class City(models.Model):
    nome_cidade = models.CharField(max_length=100)
    estado_cidade = models.CharField(max_length=100)
    pais_cidade = models.CharField(max_length=100)

    def __str__(self):
        return self.nome_cidade

    class Meta:
        permissions = (
            ('view_city', 'Can view city'),
        )
...

Thank's for help!

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