Re: Working with ModelForms

On Sat, May 3, 2014 at 9:07 AM, LaPerl wrote: > Hi all, > > I have 5 different models all with relations with themselves. I want to > use ModelForm to generate a form that will contain fields of all these > models. I read and read, I tried different things to do this

Re: I can't run django server nor activate the virtualenv

You are welcome, glad it helped :) -- You received this message because you are subscribed to the Google Groups "Django users" group. To unsubscribe from this group and stop receiving emails from it, send an email to django-users+unsubscr...@googlegroups.com. To post to this group, send email

Re: I can't run django server nor activate the virtualenv

FINALLY the server is running the problem was is two dependencies django-userena and django-bootstrap-form. and the userena dependency was relying on another dependency which is six (the core of the problem) because it didn't installed automatically when installing userena so i had to install

Re: I can't run django server nor activate the virtualenv

the boostrapform was installed correctly but there's still some problems as you said this is not my code, and i will try to contact the guy who write the code and get the missing dependencies On

Re: I can't run django server nor activate the virtualenv

Actually, its probably the same problem as from the start as I see your install from the requirements file only installed Django itself and nothing else. THerefore I presume it does not have a list of all the dependencies you need. Assuming the last error was again "ImportError : No module

Re: I can't run django server nor activate the virtualenv

Can you copy the complete error across? The end of it is all important and its chopped off in the screenshot. Thanks On Saturday, 3 May 2014 16:25:57 UTC+1, Fred DJar wrote: > > Thanks > the second solution worked fine and the requirements were installed > > but again when i tried to run the

Re: I can't run django server nor activate the virtualenv

Thanks the second solution worked fine and the requirements were installed but again when i tried to run the server this is what i got On Monday, 28 April 2014 11:03:40 UTC+1, Fred DJar wrote: >

Bug in django admin: list_editable should be not editable if appear in readonly_fields

Hello dev team! I think that if some field set to *readonly_fields* than we should display disabled field in *list_editable*. Now I have set field "*price*" into *list_editable* and *readonly_fields*. And I see this field not editable in "change view" in admin panel, but I see it editable in

Best way to use Django with Knockout.js

Hi, I was wondering if there are any good resources for demonstrating how best to use django with knockoutjs. I've come across a couple of projects which might help: http://www.django-rest-framework.org/ https://github.com/Miserlou/django-knockout-modeler I can see how I'd easily interact with

Re:

On 3/05/2014 4:41 PM, Igor Korot wrote: Hi, ALL, This is my first official post here so be gentle... ;-) Ok. Its usually worthwhile putting a subject in so people can decide whether to look at your email. The old hands wouldn't bother with a blank subject email and they are probably the ones

Re: how to break an email list into manageable sizes for emailing

This task would be better left to a background batch processor such as Celery, which integrates nicely via the Django-celery package. Submitting the form would place a job into the Celery queue, and within that job you can pull 100 (or however many you need, see

Re:

It looks like you are trying to emulate a many-to-many relationship between two models. Tables 1 and 2 would be generated by model definitions, and table 3 would be created automagically by linking the two models together via a M2M field (which would NOT add an 'id' field as you've requested

[no subject]

Hi, ALL, This is my first official post here so be gentle... ;-) I'm trying to make an application based on the following db structure: table 1: id - primary key, fname char(40), lname char(40) table 2: id - primary key, position char(20) table 3: table1_id, table2_id - foreign key Now, from