Sorry, I didn't see the django.contrib.admin. You're overrriding the add
method for you model? Because it's appear the add method is returning None:
Exception Type: TypeError at /admin/myapp/mymodel/add/
Exception Value: 'NoneType' object has no attribute '__getitem__'
2013/10/25 Derek
> Yes,
Yes, I simplified the names in the installed list and traceback, and as
result probably altered the correct grammer slighlty ... but I am hoping
that someone who understand the "innards" of the Django admin can point
what is _actually_ wrong; given that the error trace is not actually
referenci
In installed apps you have s2.myapp, the traceback is just myapp. Also you
have 's2.myapp, in the installed_apps, should it not be 's2.myapp', (ie
missing the end quote)?
Cheers, Nigel
07914 740972
On 24 October 2013 14:20, Derek wrote:
> Yes, its there - the line after 'suit'.
>
>
> On Wed
Yes, its there - the line after 'suit'.
On Wednesday, 23 October 2013 19:22:54 UTC+2, Odin wrote:
>
> Can't see the django.contrib.admin in INSTALLED_APPS.
>
>
> 2013/10/23 Derek >
>
>> I am trying to add a model entry, via the admin; I get this error (which
>> has not happened before), running D
Can't see the django.contrib.admin in INSTALLED_APPS.
2013/10/23 Derek
> I am trying to add a model entry, via the admin; I get this error (which
> has not happened before), running Django on Ubuntu 12.04 with the test
> server. Not sure if there is an error with the model (which has worked
>
I am trying to add a model entry, via the admin; I get this error (which
has not happened before), running Django on Ubuntu 12.04 with the test
server. Not sure if there is an error with the model (which has worked
fine up to now) or ...?
Request Method: POST
Request URL: http://127.0.0.1:8000/a
Also i found https://github.com/seveas/django-echelon very usefull
Specially *CurrentUserFiled*
For the information, use shoulda be always authenticated so it's not a case
here!
Thank you
On Sun, May 13, 2012 at 6:27 PM, Rafał Stożek wrote:
> It is possible using simple middleware and
> http://d
It is possible using simple middleware
and http://docs.python.org/library/threading.html#threading.local
class. But note that code doesn't have to always be called from HTTP - it
means that you must handle situation where user is not known.
Examples:
http://djangosnippets.org/snippets/2179/
htt
Hi
Let me explain it. i developing and app to trace the models activity, hmmm.
actualy it's reversioning app for other models. As a reversioning model
shoulda store user_id. What i'm trying to do is to connect those models
post_save signal to create a reversion on each changes. But i don't want
Hello all,
I am just starting learning Django and managed to get a basic website
working with a nice admin section. I am now facing a problem which I am not
sure how to resolve.
So I basically have 2 tables in my database:
- Flat
- PendingFlat
The PendingFlat would be a subset of the Flat
Duh!! yeah restarting apache did it. How did I forget something so
simple!!
Thanks
Simon
On 22 Feb, 15:04, Shawn Milochik wrote:
> Did you manually restart the Apache instance for this app?
>
> I had a similar problem, and it turned out that I had a bit of code in one of
> my model that was
Did you manually restart the Apache instance for this app?
I had a similar problem, and it turned out that I had a bit of code in one of
my model that was pulling from a table that was no longer defined. Check for
that as well.
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Hi
I'm still developing my app and I have just refactored my models in my
django app. I dropped most of the existing tables on my local dev
machine and then completely rebuilt everything using syncdb,
everything worked fine, including the admin, no probs at all.
Then I updated everything on my w
Ok, I think your answer is here on creating permissions:
http://docs.djangoproject.com/en/dev/topics/auth/#id1
in the ModelAdmin, override the `save_model` method and check the
current user permission:
request.user.has_perm('permission_name')
for more on checking permission see thi
not exactly,
for example I want to make custom permission which enables user to
change one field in model but only to 5 letter length word while other
users may change that field to 6 letters length word. Thats sily
example but in some cases it may be useful.
On Oct 6, 10:19 am, "Bogdan I. Bursu
i think you want to create a group.
On Tue, 2009-10-06 at 01:05 -0700, elminio wrote:
> Hi,
> In admin Panel I can choose groups and give them specified
> permissions. For example
>
> admin | log entry | can add log entry
>
> And I would like to create my own type of permission which for exampl
Hi,
In admin Panel I can choose groups and give them specified
permissions. For example
admin | log entry | can add log entry
And I would like to create my own type of permission which for example
is called all and gives to the specified user permission to do
whatever he likes, and other permiss
On Sep 26, 1:53 pm, bruno desthuilliers
<[EMAIL PROTECTED]> wrote:
>
> # relative import is simpler and safer
> import models
>
> # using introspection
> from django.db.models import Model
> for name in dir(models):
> obj = getattr(models, name)
> if obj is not Model and issubclass(obj,
On 26 sep, 18:07, "Lance F. Squire" <[EMAIL PROTECTED]> wrote:
> Converting a .96 site I was working on to the 1.0 set-up.
>
> Is there a quick way to list all my models in the admin?
>
> Or do I have to :
>
> from HCVGM.systems.models import System
>
> admin.site.register(System)
>
> for every on
Converting a .96 site I was working on to the 1.0 set-up.
Is there a quick way to list all my models in the admin?
Or do I have to :
from HCVGM.systems.models import System
admin.site.register(System)
for every one?
b.t.w. I presume the
class Admin:
pass
in the Models.py is
>
> However, when I try to go to the admin I get either one of two errors
> (if I keep refreshing, the errors change between the two randomly.
>
> Either
> ---
> TypeError at /admin/
>
> dict objects are unhashable
It seems that one of your errors is in books/data/admin.py and not in
the adm
fyi
I think I tracked this down.
I had this Admin def:
--
class BookVersionAdmin(admin.ModelAdmin):
fields = (
("Book", {'fields':('title', 'book', 'language', 'version',
'license', 'is_release', 'is_public')}),
("Contributors", {'fields':('authors', 'con
I am working on migrating an Django app to 1.0.
I am running into problems getting my models registered with the
admin. I am following the info here:
http://code.djangoproject.com/wiki/BackwardsIncompatibleChanges#Mergednewforms-adminintotrunk
I have updated my main urls.py:
-
from dja
Thanks Nathan. I think this solves one part of my problem. Also just
for reference, the method should be modified as so:
def company(self):
return self.sub_dept.dept.company
and so on. The exclusion of self, leads Django to look for a global
named "sub_dept".
Any idea how I can add company to
Nathan Ostgard skrev:
> list_display can use functions as their value... so, you can have:
>
> class Employee(models.Model):
> sub_dept = ForeignKey(SubDept)
>
> def company(self):
> return sub_dept.dept.company
>
> def dept(self):
> return sub_dept.dept
>
> class Admin:
> list
list_display can use functions as their value... so, you can have:
class Employee(models.Model):
sub_dept = ForeignKey(SubDept)
def company(self):
return sub_dept.dept.company
def dept(self):
return sub_dept.dept
class Admin:
list_display = ('first_name, 'last_name', 'compa
Actually my question is two fold, most likely due to brain farts on my
end. I'm relatively a newbie to Django, so please bear with me.
Here's my scenario. I have this for my relationships:
Company->Dept->Sub-Dept->Employees->Roles
"->" represents a one-to-many relationship.
Problem #1:
If I do
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