Thanks for your responses.
How I can generate multilevel lists based on fields:
category
- subcategory
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On 09/10/2011 11:55 PM, Gil wrote:
I'm a newbie to object-oriented programming too, so I suppose my
mindset is still in procedural, SQL and "recordset" modes, and
otherwise: my old ways :-) Thanks for that additional info. Good Points.
*From:* Shawn Milochik **
You're welcome. The way you wer
I'm a newbie to object-oriented programming too, so I suppose my mindset is
still in procedural, SQL and "recordset" modes, and otherwise: my old ways
:-) Thanks for that additional info. Good Points.
From: Shawn Milochik
To: django-users@googlegroups.com
Sent: Saturday, September 10, 2011 6:12
Hi nara,
Please post the debug output you get when you try to view the admin
page from your browser.
Meanwhile, while switching versions of django, you have to make sure
you remove completely every file from the old version. Failure to do
that will break django.
On 9/10/11, nara wrote:
> ok, I s
I don't know if this is the exact problem, but I've addressed this
previously here:
http://www.pineappledonut.org/2010/12/05/data-collations/
I've only worked with MySQL before, so I'm not sure about how
collations work in POSTGres, but it may provide some pointers.
cheers
L.
On Sat, Sep 10, 20
Due to an issue with yesterday's 1.2.6 release package, today we are
issuing Django 1.2.7. All users of 1.2.X Django should upgrade to
1.2.7, rather than to 1.2.6.
Details here:
https://www.djangoproject.com/weblog/2011/sep/10/127/
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"Bureaucrat Conrad, you are technically correct -- the best
On 09/10/2011 07:46 PM, gillwill2...@yahoo.com wrote:
It was only a problem insofar as I didn't like the way such a listing
looked, particulalrly when I have several models that already,
originally, have an "s" at the end of their name like "Movies",
"Books", etc... which would then be listed
It was only a problem insofar as I didn't like the way such a listing looked,
particulalrly when I have several models that already, originally, have an "s"
at the end of their name like "Movies", "Books", etc... which would then be
listed as "Moviess", "Bookss"... and removing the "s" from th
I have read https://code.djangoproject.com/wiki/CookBookThreadlocalsAndUser and
numerous other discussions about the use of threading.local() yet I still
see it being employed in various projects.
I have recently implemented a referral system in which the use of
threading.local() makes the app ver
Hey Guys,
I have a very simple stub of a project. I'm trying to do something
that should be very simple but isn't as easy as I hoped.
I have a UserProfile class. This object is basically just an extension
of the standard User Class. Within my UserProfile class, I want to
have a list of PublicProf
Actually the solution is to define a verbose_name and a verbose_name_plural
in your model's Meta
class MyModel(models.Model):
field = models.CharField(max_length=200)
class Meta:
verbose_name = 'My model's name'
verbose_name_plural = 'My model's name in plural context'
E
Dnia 2011-09-10, sob o godzinie 14:50 -0700, Christian Ramsey pisze:
> def __unicode__(self)
> return 'Name you'd like without the s'
> for each model and this will be used instead.
That's obviously not what he's after.
> On 10 Sep 2011, at 14:40, Gillwill wrote:
> >Apparently the Django
y0
I have just published an application that serves as a social platform
to share stuff to anyone in channels. It has one single textbox, it's
command driven, it's completely ajax based, and is open for a lot of
modifications and improvements.
You can check it out here www.gsyko.com
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You rec
ok, I switched to the 1.3.1 released version of Django, and
also tried Python2.6 instead of Python2.7 on the dev version
of Django. Things are still majorly broken, see the following
in the django shell on a fresh startproject, I could not
import even the top level django module. Then,
I set PYTHON
I believe you can use :
def __unicode__(self)
return 'Name you'd like without the s'
for each model and this will be used instead.
On 10 Sep 2011, at 14:40, Gillwill wrote:
> Apparently the Django default is to append the letter "s" to the end
> of the model name for each listed under
Apparently the Django default is to append the letter "s" to the end
of the model name for each listed under a given application on the
Site Administration page. (It does this in the tutorial sample site as
well - e.g. naming "poll" "polls", etc...)
Is there any way to get rid of that?
I would th
I'll get through this yet :)
I tried the commands you have under the django shell, and I got
'example.com'
on the django.Site query, and not an error. Also, interestingly,
within the shell, I don't see
django on the sys.path at all (shown below), but I do see my project
mblog. This could
be the ca
Try this one for 1 question: {{object.get_FIELDNAME_display}} - for you it
will be look like this {{ c.get_category_display }}
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Also look here:
https://docs.djangoproject.com/en/1.3/ref/models/instances/#django.db.models.Model.get_FOO_display
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It seems I am getting an error from the WSGI server after all. As a reminder,
my uwsgi process is started from upstart like this:
exec python /var/www/NurseTriage/triagedb/manage.py runfcgi protocol=scgi
method=threaded host=127.0.0.1 port = 3033
When I use Firefox to browse to 127.0.0.1:3033,
maybe try:
{%for c in category.elements %}
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2011/9/10 Danny Gale
> Daniel, thank you very much. That's extremely helpful.
>
> What I'm trying to define is that every song can be on one album, but each
> album can (of course) have many songs. For that, would it be better to use
> the ManyToMany from Album to song or FK from song to album?
>
In my model I use: choices=CATEGORY_CHOICES property.
CATEGORY_CHOICES = (
('hw', "Hardware"),
('soft', "Software"),
('snd', 'Audio'),
)
In templatetags it looks like this:
{%for c in category %}
Well, I am willing to go to the Big Nerd Ranch or other hard core fast pace
Django
learning course. I have a excellent understanding of programming and will
excel very
quickly. I am willing to invest in your future if you are willing to invest
in mine.
I would love to talk about my background and g
On Sat, Sep 10, 2011 at 8:36 PM, Danny Gale wrote:
> Daniel, thank you very much. That's extremely helpful.
>
> What I'm trying to define is that every song can be on one album, but each
> album can (of course) have many songs. For that, would it be better to use
> the ManyToMany from Album to son
Daniel, thank you very much. That's extremely helpful.
What I'm trying to define is that every song can be on one album, but each
album can (of course) have many songs. For that, would it be better to use
the ManyToMany from Album to song or FK from song to album?
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Following the thread, we at Blackchair Software are also looking for django
developers in the near future so if anyone would like to send their resume.
Regards,
Carlos Daniel Ruvalcaba
Blackchair Software
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Just to follow up on this, Simplicity Media is also looking to take on
further Python/Django developers in the near future. We've got the point
where we are having to turn down work, so if you are interested, please send
across your portfolio/resume/linkedin etc.
Thanks
Cal
On Sat, Sep 10, 2011
Hey there gang. If any of you are in need of employment, there
is some quality job positions available with good pay for any decent
Django programmers. The consulting firm is based out of St. Paul
Minnesota and can allow you to work from home. The web site is
http://www.comfychairconsulting.com/
On
On Saturday, 10 September 2011 05:46:33 UTC+1, Danny wrote:
>
> Hi, I'm learning Django (albeit slowly) and I'm trying to set up a really
> simple database. I want to have Artists, Albums, and Tracks. I want to be
> able to navigate both ways in the db. So you should be able to go from
> artis
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