You have a comma at the end of this:

        bar = models.IntegerField(),

Could that be the issue?

François

On Apr 3, 2014, at 6:24 PM, Jorge Arevalo <jorgearev...@libregis.org> wrote:

> Hello,
> 
> I have this dummy models.py
> 
> from django.db import models
> 
> class Foo(models.Model):
>     bar = models.IntegerField(),
>     // more integer fields here...
>     xyz = models.IntegerField()
> 
> And I want to know the SQL code generated for PostgreSQL, executing:
> 
> python manage.py sqlall myapp
> 
> No south involved. Just Django 1.5.4. And this is the SQL output
> 
> BEGIN;
> CREATE TABLE "myapp_foo" (
>     "id" serial NOT NULL PRIMARY KEY,
>     "xyz" integer NOT NULL
> )
> ;
> 
> COMMIT;
> 
> Just the internal id and the last specified field were generated. The rest 
> are ignored. No matter how many fields or what fields. Just takes the last 
> one, and generates that output.
> 
> Does it make any sense? I'm running Django 1.5.4 under a virtualenv, using 
> Python 2.7 in Ubuntu 12.04. This is the complete output of pip freeze in my 
> virtualenv
> 
> Django==1.5.4
> Jinja2==2.7.1
> MarkupSafe==0.18
> PIL==1.1.7
> Pygments==1.6
> Sphinx==1.1.3
> Unipath==1.0
> argparse==1.2.1
> distribute==0.6.24
> django-admin-tools==0.5.1
> django-appconf==0.6
> django-bootstrap3==2.0.0
> django-colorful==0.1.3
> django-compressor==1.3
> django-extensions==1.2.5
> django-geojson==2.1.1
> django-guardian==1.1.1
> django-leaflet==0.8.2
> django-model-utils==1.5.0
> django-secure==1.0
> django-sorting==0.1
> django-waffle==0.9.2
> docutils==0.11
> feedparser==5.1.3
> psycopg2==2.5.1
> requests==2.0.1
> six==1.4.1
> wsgiref==0.1.2
> 
> This is just... weird. Any clues?
> 
> Many thanks in advance
> 
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