Allow custom path to wsgi application

[k4ml]

Currently the generated .wsgi script pull out the application from django.core.handlers.wsgi.WSGIHandler(). Sometimes it's desirable to specify custom application callable, such as when integrating with third party debug application. Ideally, with settings such as:-

[django]
...
wsgiapp = myproject.wsgi.application

The generated .wsgi script will import the application as is, rather than calling djangorecipe.wsgi.main() function.

I've worked a bit on this but so far I have to follow djangorecipe approach of calling something like myproject.wsgi.main(). Since default django template already include myproject.wsgi module with attribute named application, it would be much better if we can follow the same pattern.

[jellegerbrandy]

Wouldn't settings.WSGI_APPLICATION be the place to specify the wsgi app?

https://docs.djangoproject.com/en/dev/ref/settings/#std:setting-WSGI_APPLICATION

[k4ml]

That I think does not allow you to fully wrap the application, such as even before django itself get called.

[jellegerbrandy]

I'm not really sure what you mean by "wrapping the application before django gets called" - how would that work exactly?

In any case, as far as i can see, having the wsgi script respect the WSGI_APPLICATION setting of django (just as the "runserver" command does) is something that seems a good idea in any case...

[reinout]

I've merged #78, I guess that solves the issue in a generic way? If needed you can wrap/modify the wsgi application object that way.

Can you take a look at djangorecipe 1.9 and its deploy_script_extra option?