You are probably getting this error because when you generate your list
from categories. you are adding sub categories as list. so your list will
be like following
[cat1,cat2,cat3,[subcat1,subcat2],[subcat3,subcat4]]
I think you want to use
list.extend(BuildList(category.childrens.all()))
so y
For development, you could add this to main urls.py like this.
(r'^static/(?P.*)$', 'django.views.static.serve',
{'document_root': settings.STATIC_ROOT, 'show_indexes':True}),
But for production use your web server to serve static files.
At least this is how I solved it. I hope that helps.
O
This is why linux is the best os. If anything goes wrong, you can
blame it on linux :)
1) make sure you are really using django1.4a in server (from django
import VERSION). maybe you installed multiple versions of django and
using the wrong one (you might have 1.2.x on system and 1.4a on
virtualenv
hat design mistake. but I don't get this error in zope version.
On Jan 4, 2:49 pm, Thomas Weholt wrote:
> He probably meant zope.interface :-)
>
> Regards,
> Thomas
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> On Wed, Jan 4, 2012 at 1:47 PM, huseyin yilmaz
> wrote:
> &g
Could you direct me to an example (or documentation). I could not find
any source about zone.interface.
On Jan 4, 2:17 pm, Donald Stufft wrote:
> Why not use zone.interface
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> On Wednesday, January 4, 2012 at 7:07 AM, huseyin yilmaz wrote:
> > Hi eve
Hi everybody,
I want to implement interface functionality in python. I wrote
following sample code
https://gist.github.com/1559689
Here I use an abc as an interface.
Couple of my models are implementing this interface and one of my
consumer model is using those models. In this implementation, I
Generic forign key might help you
https://docs.djangoproject.com/en/dev/ref/contrib/contenttypes/#django.contrib.contenttypes.generic.GenericForeignKey
You might not even need a metadata here.(The metadata you need looks
like ContentType model)
Just do
class GenericManyToManyDetail(Model)
ct1 = .
It looks to me like you do not have 500.html file in your base
template directory. If you have a server error in production mode
(DEBUG = False), django will show 500.html to users. So users will not
see actual error message. But you will get that message by email.
Since you do not have 500.html it
I was going to say emacs, but since you also use windows, Aptana would
be a better fit.
On Dec 19, 12:34 pm, Alec Taylor wrote:
> I'm looking for a Django IDE which incorporates the following features:
> - Syntax-highlighting
> - Projects (all code file of the project shown separated by director
With select related, hole queries can be ommitted but this time for
every score object one hole object will be created. so there will be
alot of hole objects.
On Dec 5, 12:18 pm, Håkon Erichsen wrote:
> 2011/12/5 kenneth gonsalves
>
> > The code looks simple to me, but to analyse 10,000 rounds
First of all You are queryin everything that you have on database for
this view and making django model objects from those. For a large db
this will run very slow and for larger dbs, it will throw memmory
exception.
Try something like this:
for hole in Hole.objects.filter(tee_course=club):
...
Also check memcached configuration (/etc/memcached.conf)
from this file,you can limit IP addresses that memcached listening to
here is a configuration parameter that you might have
# Specify which IP address to listen on. The default is to listen on
all IP
addresses
# This parameter is one of the
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