CODEWITHSUNDEEP
djangomodelpermissions
A.Raouf
A.Raouf

Reputation: 2320

Custom creation of permissions in django model

Now I have models.py like this

class FileCategory(models.Model):
    file_type = models.CharField(_('type'), max_length=128)

    def __unicode__(self):
        return self.file_type


class Doc(models.Model):
    owner = models.ForeignKey(settings.AUTH_USER_MODEL,
        on_delete=models.CASCADE)
    file_category = models.ForeignKey('FileCategory', null=True)
    file = models.FileField(upload_to='documents/',
                            null=True, blank=True)

    class Meta:
        permissions = (("upload_file", "upload file"),
                       ("delete_file", "delete file"),
                       ("download_file", "download file"),
                       ("access_file", "access file"),
                       )

Now I want to add permissions for every FileCategory table For example If I have filecategory numm, dumm and summ I want to automatically have permissions ("access_file_numm","access_file_summ","access_file_dumm", )

I tried to make a solution like this:

permissions = (bla,bla,bla,
                   )+\
                  ("access_file_%s" (type.__str__() for type in FileCategory.objects.all()),
                   "access %s"(type.__str__() for type in FileCategory.objects.all()))

And sure I want to migrate them to the database

Solution:

as Moemn answers with an edit:

def save(self, *args, **kwargs):
    super(FileCategory, self).save(*args, **kwargs)

    Permission.objects.get_or_create(content_type=ContentType.objects.get_for_model(Doc),
                              codename="access_type_%s" % self.file_type,
                              name="access type %s" % self.file_type)

Upvotes: 1

Views: 118

Answers (1)

Moamen
Moamen

Reputation: 706

You can do this by creating Permission dynamically after every FileCategory creation.

e.g.

def save(self, *args, **kwargs):
    super(FileCategory, self).save(*args, **kwargs)
    Permission.objects.create(content_type=ContentType.get_for_model(Doc), name="access_file_%s" % self.file_type)
    Permission.objects.create(content_type=ContentType.get_for_model(Doc), name="access_%s" % self.file_type)

Or you can use signals, the same idea.

Upvotes: 1

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