Hi Ben,
That is interesting approach. So I can get out of using app globals. I
will test and update the RUM community on this.
Can I use decorators for authorization here? I have repoze.who & what
in place.
RumController = RumApp({
................
...............
})
Thanks,
Krish
On Dec 21, 7:16 am, Ben Bangert <[email protected]> wrote:
> On Dec 20, 2009, at 5:49 PM, [email protected] wrote:
>
> > I use app_globlas to create and store the RUM based wsgi application.
>
> Odd, I used RUM in an app, it's easier than that actually.
>
> Make your RumApp instance inside a controller (say, called rum.py). If you
> make it inside of 'rum.py' then name the instantiated app instance
> RumController.
>
> Ie. do your:
>
> RumController = RumApp({
> 'debug': asbool(config.get('debug')),
> 'full_stack': True,
> 'rum.repositoryfactory': {
> 'use': 'sqlalchemy',
> 'session_factory': sf.model.meta.Session,
> 'models': [
> sf.model.Page, sf.model.PageItem,
> sf.model.AdminPanel, sf.model.AdminTask
> ]
> },
> 'rum.viewfactory': {
> 'use': 'toscawidgets',
> },
> 'rum.translator': {
> 'use': 'default',
> 'locales': ['es', 'en'],
> }
> })
>
> Then setup a route to go to it:
> map.connect('/rum/*path_info', controller='rum')
>
> That should be it. This utilizes the fact that all Pylons controllers are
> technically WSGI app callable, so as long as you follow the naming
> convention, it doesn't *have* to be a Pylons WSGIController (note this is
> what you inherit from in your lib/base.py), it can be *any* WSGI app.
>
> Cheers,
> Ben
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