Django - how to create a private subpage?

[Petey]

I am trying to do a private subpage for every user which will be accessed trough url:
    url(r'^galeria/$', 'publisher.views.private_gallery'),

I was trying to create a query set for private gallery by using a signal: user_logged_in

p = Publisher.objects.filter(status='p', pub_type='private_gallery', user_gallery=str(user_logged_in)).order_by('-id')

However, django does not accept signals in querysets.
Could you give me some good hints on making private pages which can be viewed only by Users?

[Kurtis Mullins]

Check out this page: https://docs.djangoproject.com/en/dev/topics/auth/#handling-object-permissions
Also, you may want to check out decorators if the user only has to be logged in. Another method you could use are groups.

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[Victor Hooi]

heya,

I might be misunderstanding your requriements, but could you use the @user_passes_test decorator with a has_perm check?

@user_passes_test(lambda u: u.has_perm('private_pages.foo'))

https://docs.djangoproject.com/en/dev/topics/auth/

You can probably make the lambda a bit smarter, instead of using has_perm, check if the pagename matches the username.

Cheers,

Victor

[Matt Schinckel]

That line (p = ...) is in a view, right?

It will probably want to be:

    p = Publisher.objects.filter(status='p', pub_type='private_gallery', user_gallery=request.user).order_by('-id')

Matt.