Django internationalization language codes
Where can I find list of languages and language_code like this.
If you want something you can use from within django, try:
from django.conf import settings
this will be in the format above, making it perfect for assignment in one of your models choices= fields. (i.e. user_language = models.CharField(max_length=7, choices=settings.LANGUAGES))
LANGUAGES = ( ('ar', gettext_noop('Arabic')), ('bg', gettext_noop('Bulgarian')), ('bn', gettext_noop('Bengali')), etc.... )
Note about using settings:
Previous answers mention only getting LANGUAGE from settings.py, hovewer there is a big chance that this variable will be overwritten. So, you can get the full list from django.conf.global_settings.LANGUAGES
from django.db import models from django.conf.global_settings import LANGUAGES class ModelWithLanguage(models.Model): language = models.CharField(max_length=7, choices=LANGUAGES)
from django.conf import settings #note settings is an object , hence you cannot import its contents settings.configure() #note LANGUAGES is a tuple of tuples lang_dict = dict(settings.LANGUAGES) #use lang_dict for your query. print lang_dict['en']
I understood from Django Project that you can only use a dummy gettext function:
If you define a custom LANGUAGES setting, as explained in the previous bullet, it's OK to mark the languages as translation strings -- but use a "dummy" ugettext() function, not the one in django.utils.translation. You should never import django.utils.translation from within your settings file, because that module in itself depends on the settings, and that would cause a circular import.".
It took me some time to find the solution, but I finally got it; the choices of the model field needs to have a tuple with real gettext functions, with a lambda function the dummy's can be wrapped in the real gettext functions as follows:
from django.utils.translation import ugettext_lazy as _ language = models.CharField(max_length=5, choices=map(lambda (k,v): (k, _(v)), settings.LANGUAGES), verbose_name=_('language'))