Question

I started to play with new i18n_patterns in Django 1.4. Basically, i want to have language links for each of my supported languages on all of my templates headers. I have implemented my header as a separate template that is being included in other templates.

Is there a way to keep my header generic and solve this without passing the current view name or current url in template context? I guess it comes to a question how do i retrieve the current view or url from inside the template in a generic way.

BTW, i discovered that my previous approach with set_lang view to change the active language using the referrer will be broken with url_patterns as after changing the language it will change it back when redirected to the referred view.

Any help figuring out the common approach to set language links in templates to be used with url_patterns in a generic way would be appreciated!

Answer 1

My apologies for the long delay. Thank you all for your answers.

First of all to comment on the two solution options by Chris:
Neither custom set_language nor javascript are good for the purpose as the whole beauty of url patterns is being SEO friendly.

Furthermore, simply replacing the prefix language in URL cannot be treated as full solution for urlpattern based urls as the whole URL might be translatable too. Ex: /en/profile/ for english and /fr/profil/ for french.
To solve such a problem one needs to capture the viewfunc and the arguments in order to reverse it for different language.

Fortunately for me, my project does not use translatable URLs for now and I took the following approach.

1. Add django.core.context_processors.request to TEMPLATE_CONTEXT_PROCESSORS so that request to be available in template rendering context.

2. Use RequestContext in views when rendering your views. This is always the case for me independent of the topic.

3. Write a quite simple templatetag that requires the context and takes an argument a language code to return the current URL in the given language. It basically makes sure the current URL has valid language-code prefix and simply returns another string that is the same current URL with replaced language-code.

ex:

from django import template
register = template.Library()

@register.simple_tag(takes_context=True)
def get_current_url_for_lang(context, lang_code):
    request=context.get('request',False)
    if not request:
        return None

    request=context['request']
    curr_url=request.path
    if len(curr_url) < 4 or curr_url[0] != '/' or curr_url[3] != '/':
        return curr_url

    if context.get('LANGUAGES',False):
        codes = []
        for code,name in context['LANGUAGES']:
            codes.append(code)

        curr_langcode = curr_url[1:3]
        if lang_code not in codes or curr_langcode not in codes:
            return curr_url

    changed_url = '/'+lang_code+curr_url[3:]
    return changed_url

Furthermore, if one does not like to inject full request into context it will be quite straightforward to write your own context_processor that simply pushes the request.path as current_url_path for instance and use that instead in your templatetag.

Your comments are welcomed as always!