On Sun, Jan 31, 2010 at 11:46 PM, esatterwh...@wi.rr.com <
esatterwh...@wi.rr.com> wrote:
> if if you want to change the menu based on the user, you could
> probably just use the user permissions from the auth context processor
>
> if the user has the permissions ( access ) to the option - show it
>
> else - don't show it.
>
>
> or if you want to use the model you have listed here, you could use
> the {% ifequal tag %}
>
> {% ifequal myuser.access 1000 %}
> show some stuff...
> {% else %}
> show something else...
> {% endifequal %}
>
The problem with that is if the user has permission 1000 for superuser and
there is a menu item for staff requiring a permission of 100 the if check
would fail. My programmers brain says, no problem just use a bit vector, but
django templates don't allow that either. In the development version it's
possible to do;
{% if myuser.access >= 100 %}
But i'm looking at deploying the site so i don't really want to use it.
>
> I personally think using putting users in groups and assigning
> permissions then checking in the template for who has what would be
> the easiest solution. Django does most of the work for you
>
That may be a better option, i'm still learning django and may have cheated
with access permissions. I'll look into it but i want to avoid a mess of if
checks. Since the menu is dynamically generated i can't assume a menu item
with any particular permission is going to arrive.
The solution i did use seems to be working quite well. In the template the
"access" tag works like this.
{% for item in menu.objects.all %}
{% access item %}
#Do stuff
{% endaccess %}
{% endfor %}
The "access" tag checks for the existence of user in the context and
calculates the users access level, or failing that assumes an access of 1
(Anonymous). Then it checks for an access property on item and renders the
block only if user.access >= item.access . You see why the formerly
mentioned if check is way simpler?
> On Jan 30, 9:28 pm, Dylan Evans <dy...@contentfree.info> wrote:
> > No i have processors, i have all the data i need in the template, i just
> > don't know an elegant way to condition out menu items.
> >
> > This is how my model works
> >
> > ACCESS_CHOICES = (
> > (1, "Public"),
> > (10, "Private"),
> > (100, "Staff"),
> > (1000, "System")
> > )
> >
> > class Menu(models.Model):
> > name = models.CharField(max_length=16)
> >
> > class MenuItem(models.Model):
> > menu = models.ForeignKey(Menu)
> > access = models.IntegerField(choices=ACCESS_CHOICES)
> >
> > class MenuSubItem(models.Model):
> > item = models.ForeignKey(MenuItem)
> > access = models.ForeignKey(MenuSubItem)
> >
> >
> >
> >
> >
> > On Sun, Jan 31, 2010 at 5:31 AM, Shawn Milochik <sh...@milochik.com>
> wrote:
> > > I think you've missed context processors, which is easy to do. I'm
> assuming
> > > that your issue is that you want to have something passed in the
> context on
> > > every page load to do something like decide which menu items are
> available
> > > based upon whether the user is logged in, their privileges, or
> whatever.
> >
> > > Context processors allow you to define a dictionary that gets appended
> to
> > > the context of every response you send, so you can have that common
> stuff
> > > there.
> >
> > > 1. Write code to get the values appropriate for the current user or
> > > whatever. Put these in a Python file in your app.
> > > 2. Add that file to the TEMPLATE_CONTEXT_PROCESSORS in your
> settings.py.
> > > 3. Replace Context() with RequestContext() in your render_to_response
> > > calls.
> >
> > > If I've missed your actual point, please give more detail. I think this
> > > simplifies what you're trying to do.
> >
> > > Shawn
> >
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> > "The UNIX system has a command, nice ... in order to be nice to the other
> > users. Nobody ever uses it." - Andrew S. Tanenbaum
>
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