You would either have to duplicate the Title foreign key in the Copy
model or write a custom check in the save method.
To enforce uniqueness over several columns you would use the
"unique_together" field of the Meta class for your Copy model. (see:
http://www.djangoproject.com/documentation/model
hatever').focus(someJavascriptFunction);
});
As for classes, I've taken to applying them to a div surrounding my
input elements. So I have something like:
{{ form.whatever.label }}
{{ form.whatever }}
The CSS isn't much longer...
form .red input { color: red; }
Nathan Ostgard
On Jul 8, 8:04
Do this instead:
PreventivoForm = form_for_model(Preventivo)
del PreventivoForm.base_fields['codice']
f = PreventivoForm()
f2 = PreventivoForm(auto_id='id_%s2')
Nathan Ostgard
On Jul 9, 2:33 am, ilDave <[EMAIL PROTECTED]> wrote:
> Hi!
> I have to show two
and flexible
for styling:
{{ form.some_field.label }}:
{{ form.some_field }}
{{ form.some_field.errors }}
... more fields ...
... more fieldsets ...
or cancel
Nathan Ostgard
--~--~-~--~~~---~--~~
You received this message because yo
Note that this will return the primary key of an instrument for the
value, rather than the name.
----
Nathan Ostgard
On Jul 10, 4:41 am, AnaReis <[EMAIL PROTECTED]> wrote:
> Hi,
> I have a form on my project which has a drop down list with several
> options that are loaded from a t
ForeignKey fields use the Field.get_choices() method to build a list
of options. Perhaps you could subclass the ForeignKey field and
customize the method?
On Jul 10, 10:30 am, Carl Karsten <[EMAIL PROTECTED]> wrote:
> I am using the django User like so:
>
> # models.py
> from django.contrib.auth.
On Jul 10, 2:58 pm, novice <[EMAIL PROTECTED]> wrote:
> but I get error that the offers_offerprice.offerseller_id column
> doesnt exist.
I think you're missing an underscore. Your model has the field named
offer_seller... shouldn't it then be
offers_offerprice.offer_seller_id?
--~--~-~-
I really think you should define a custom field and override the
choices. For example:
from django.utils.encoding import smart_unicode
class UserForeignKey(models.ForeignKey):
def get_choices(self, include_blank=True,
blank_choice=models.BLANK_CHOICE_DASH):
"Returns a list of tuples used as
It's the default behavior of Python:
>>> str(None)
'None'
I don't agree that it should be the default. None and "" are two
distinct values -- especially with a database. In databases, None (or
NULL) normally represents a _missing_ value, whereas a "" is one that
was intentionally specified to be
If you're using an SVN copy (not 0.96), there is already a context
processor for this. Add 'django.core.context_processors.media' to
TEMPLATE_CONTEXT_PROCESSORS in settings.py and you will be able to use
{{ MEDIA_URL }} in your templates.
On Jul 14, 6:43 am, skam <[EMAIL PROTECTED]> wrote:
> It
It's a question of value to the customer. Of course they don't
necessarily care about Django -- they care about the product. It's up
to you to illustrate to them, using Django, that you can deliver a
better product to them than the competition, and that the costs are
worth it. I've found that the
3. Actually setting the environment variable DJANGO_SETTINGS_MODULE in
your shell. If you don't know how to do that already, you should
probably use one of the other two options.
Nathan Ostgard
On Jul 17, 2:43 pm, Collin Grady <[EMAIL PROTECTED]> wrote:
> If you don't use &quo
You can also refer to this page in the documentation for more
information:
http://www.djangoproject.com/documentation/static_files/
Nathan Ostgard
On Jul 17, 11:35 am, Dmitriy Sodrianov <[EMAIL PROTECTED]> wrote:
> Hello everyone!
>
> Let me ask for help with the fo
raise forms.ValidationError('some error message')
MyForm = forms.form_for_model(MyModel, form=MyBaseForm)
----
Nathan Ostgard
On Jul 18, 7:06 am, stereoit <[EMAIL PROTECTED]> wrote:
> Hi,
> I have model with field that has attribute unique set to True. I know
> new
rror('some error message')
MyForm = forms.form_for_model(MyModel, form=MyBaseForm)
----
Nathan Ostgard
On Jul 18, 7:06 am, stereoit <[EMAIL PROTECTED]> wrote:
> Hi,
> I have model with field that has attribute unique set to True. I know
> newforms are not able to han
Here's something on djangosnippets.org for displaying thumbnails in
the admin list:
http://www.djangosnippets.org/snippets/239/
Or do you mean on the page for editing a record?
Nathan Ostgard
On Jul 18, 2:11 pm, Naco <[EMAIL PROTECTED]> wrote:
> Does anybody know
)$ dispatch.fcgi/$1 [QSA,L]
In my case, I have my Django projects/apps installed under ~/django,
and my Virtual Python installation in ~/python.
Nathan Ostgard
On Jul 18, 9:54 am, walterbyrd <[EMAIL PROTECTED]> wrote:
> I am trying to follow jeff croft's tutorial for inst
If you just want to specify the urls as listed, you could always
create a urls folder in myapp, put an __init__.py in there, and create
myapp/urls/feature1.py and myapp/urls/feature2.py.
Nathan Ostgard
On Jul 18, 7:37 pm, Collin Grady <[EMAIL PROTECTED]> wrote:
> You'll
manage.py syncdb won't add new fields to the database if the table
already exists. You'll have to do this yourself. See manage.py sqlall
for some help.
Otherwise, if you have: user = models.ForeignKey(User), then look for
a field named user_id in your table.
----
Nathan Ostgard
You may find it easier to avoid these conflicts by doing:
from django.contrib import auth
Then use:
user = auth.authenticate(username='foo', password='bar')
auth.login(request, user)
On Jul 19, 1:02 am, james_027 <[EMAIL PROTECTED]> wrote:
> hi,
>
> i've been trying to get started with django an
You can define a custom decorator instead, specifying a custom
login_url:
from django.contrib.auth.decorators import user_passes_test
my_login_decorator = user_passes_test(lambda u: u.is_authenticated(),
login_url='/my/login/url')
Then you can use:
@my_login_decorator
def someview(request):
.
Use forms.ModelChoiceField instead:
immagine =
forms.ModelChoiceField(queryset=Pics.objects.all().order_by('immagine'),
cache_choices=False, label='Pic', required=False)
On Jul 20, 1:15 am, ilDave <[EMAIL PROTECTED]> wrote:
> I've a form that handles file uploading to the database: it works
> fi
Python also allows you to use ** to unpack a dict into a set of
kwargs:
b = Book(**bf.cleaned_data)
On Jul 20, 4:28 am, LaundroMat <[EMAIL PROTECTED]> wrote:
> On 20 jul, 10:53, james_027 <[EMAIL PROTECTED]> wrote:
>
>
>
> > hi,
>
> > is there a short cut for assigning input data to model's data
Assuming cwd is the projects folder:
settings.py:
import os.path
MEDIA_ROOT = os.abspath('./media')
urls.py:
from django.conf import settings
urlpatterns = patterns('',
(r'^media/(?P.*)$', 'django.views.static.serve',
{'document_root': settings.MEDIA_ROOT}),
)
On Jul 20, 6:13 am, cjl <[EMA
Interesting! Didn't know document_root would figure out relative
paths. Glad you got it working.
Re. media -- yeah, sorry about that, I've taken to renaming the
admin's to media_admin.
On Jul 20, 12:26 pm, cjl <[EMAIL PROTECTED]> wrote:
> Nathan:
>
> Thank you for your reply.
>
> Your solution s
Regarding the first question, you can do this by creating a middleware
class which checks for authenticated users by default. You could also
created a decorator to tag view functions as public. For example...
in file yourproject/yourapp/middleware.py:
from django.conf import settings
from dj
Pretty sweet, good job!
I would recommend echoing some info when one runs start.bat. I thought
it wasn't working right at first because it just opened a normal
command prompt. Just telling people a few of the Django commands that
are available to them would probably be enough.
On Jul 23, 8:14 am
First, I've recently setup a Dreamhost account myself with no problems
-- feel free to email me directly if you want some help, and I'll see
what I can do.
That said...
Django and Python are still going through growing pains in the shared
hosting department. The same thing was going on with Ruby
You can specify it upon form creation. Here is an example of how you
could use prefixes for multiple model forms:
In your views:
from django import newforms as forms
from django.http import HttpResponseRedirect
from django.shortcuts import render_to_response
from yourproject.yourapp.models impor
As the error says, you're using self.question in the Choice model --
it should be self.choice.
e.g., change this:
class Choice(models.Model):
# ...
def __unicode__(self):
return self.question
to this:
class Choice(models.Model):
# ...
def __unicode__(self):
return self.choice
What import are you aliasing as _?
You have three options for gettext calls:
- ugettext, which translates as soon as it's *called*.
- ugettext_lazy, which does not translate until the attribute is
*accessed*.
- ugettext_noop, which translates at the last possible moment, such as
when the value is
You shouldn't be capturing the args yourself -- the syndication
framework will automatically split up anything after the feed name and
pass it to get_object on your Feed class. So, if "/feeds/latest/foo/
bar/" was requested, it would call your feed class with the bits
['foo', 'bar']. You can then
You can define a list of valid keys to look for:
keys = ['one', 'two', 'three', 'four', 'five']
for key in keys:
if not key in request.POST:
continue
.. do stuff ..
Or you can delete the keys you don't want before looping:
for key in ['x', 'y', 'submit']:
del request.POST[key]
for key
request.POST is a dictionary, not a list, so a for loop like yours is
iterating over the keys of the dict. to get the values, you can do:
for value in request.POST.itervalues():
to iterate over both keys and values:
for a, value in request.POST.iteritems():
On Jul 27, 2:54 pm, Greg <[EMAIL PRO
ifequal can't do filters, unfortunately.
But you can do this:
{% for item in menu_items %}
{% if forloop.last%} class="last-item"{% endif %}
{% endfor %}
Or
class="{{ forloop.first|yesno:"first," }} {{ forloop.last|
yesno:"last," }}"
On Jul 27, 11:42 pm, Eloff <[EMAIL PROTECTED]> wrote:
> T
To illustrate with the Python shell:
>>> 0 == "0"
False
>>> 0 == int("0")
True
On Jul 27, 11:10 pm, Sean Perry <[EMAIL PROTECTED]> wrote:
> On Jul 27, 2007, at 10:36 PM, Greg wrote:
>
> > AssertionError at /rugs/cart/addpad/
> > [u'0']
>
> > Does that mean that the value is 0? below is my view
I could be wrong, but I think the problem he's having is having to
specify the menu variable in the context for every view, not the
template end of it...
To fix that, you will want to check out context processors:
http://www.djangoproject.com/documentation/templates_python/#subclassing-context-re
Try: {{ data|pprint }}
On Jul 29, 10:42 pm, "[EMAIL PROTECTED]" <[EMAIL PROTECTED]> wrote:
> I'm coming from CakePHP and I would typically set a variable for use
> in my view with a call to findAll. Since there is a lot of data in the
> array, I typically do something like:
>
>
>
>
>
> This wa
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
You can use ModelMultipleChoiceField for this:
items =
newforms.ModelMultipleChoiceField(queryset=Items.objects.all())
On Aug 2, 1:24 am, james_027 <[EMAIL PROTECTED]> wrote:
> hi
>
> somewhere on the django documentation I saw the choice provided for
> MultipleSelectField is from a Database but
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