Hi,
Since it is static (i.e. never modified), the CSS classes dictionary (statemap)
could be extracted from the function and promoted as a module global variable,
to avoid it being reconstructed each time the filter is invoked.
This will not cut the processing time in half, but small streams make big
rivers :)
Eric
________________________________
From: django-users@googlegroups.com <django-users@googlegroups.com> on behalf
of treyd <t...@treyd.us>
Sent: Saturday, November 11, 2017 20:49
To: Django users
Subject: Re: Django Templates and Conditional CSS classes
Omar (and everybody else who responded),
Thanks for that ideas and putting a method on the model to return the
appropriate CSS class based on the state. I think that would be the cleanest
solution if I am going to bite the bullet and put this template-specific logic
on the model.
It got me thinking, though, and it might be a better place to handle this in
the view. I tried that and I got it to work by extending DetailView into a
custom view modified the context before rendering the template. This removed
the template logic, but I ran into another issue when I am trying to access
other models associated with the model I'm looking at in the DetailView and
handling their state label styling, so I scrapped that approach.
I did some more research and the solution I finalized on was a custom template
filter that would read the state and convert it to the appropriate CSS class,
e,g,:
<span class="label {{ myobject.state|state_css_class }}">{{ myobject.state
}}</span>
To do this, I followed the Django Custom Template Tags
How-To<https://docs.djangoproject.com/en/1.11/howto/custom-template-tags/> and
came up with this:
from django import template
register = template.Library()
@register.filter
def state_css_class(value):
"""returns appropriate bootstrap label class for states"""
statemap = {
'Scheduled': 'label-default',
'Provisioning': 'label-primary',
'Active': 'label-success',
'Deleting': 'label-primary',
'Ended': 'label-info',
'Unprovisioned': 'label-default',
'Building': 'label-primary',
'Deleted': 'label-primary',
'Error': 'label-danger'
}
try:
return statemap[value]
except KeyError:
return 'label-default'
I like this solution the best because it keeps template logic in/near the
template layer, I can add new state renderings easily, and I can extend it to
handle states for multiple model types (i have more than one stateful model in
my project).
I hope someone else finds this useful! Thanks again all for the discussion.
On Saturday, November 11, 2017 at 10:18:25 AM UTC-5, Omar Helal wrote:
Hi treyd,
I think idea of updating the model and storing it in the db is a bit overkill,
however you could create an @propery on the model that will do that logic for
you and return the css class for you.
e.g.
@property
def css_class(self):
if self.state is SUCHANDSUCH:
return "label-soandso"
and so in the template you can do what Vijay recommended but without storing
anything you don't need in the database.
Good luck with going Angular, it plays well with Django!
On Saturday, 11 November 2017 09:50:38 UTC+11, treyd wrote:
Thanks, all, for the info. Looks like the best way to clean this up is to make
the model have some knowledge of the CSS class I want to use and make a method
to get it. I like that, but I was unsure about sticking view information into
the model. Probably the best solution is javascript logic in the browser,
which I'll get to.
On Friday, November 10, 2017 at 5:45:02 PM UTC-5, Vijay Khemlani wrote:
You can also add a "state_css_class" (or something) method to your object class
and just call
<span class="label {{ myobject.state_css_class }}">...</span>
the method should be quite simple
On Fri, Nov 10, 2017 at 7:01 PM, Adam Simon <add.si...@gmail.com> wrote:
You can pass the class from either the model or the view into the template.
For example, you can assign a class to a button with a variable:
<button class="{{ class }}"><|button>
And then the class would have to be defined in your style somewhere
On Fri, Nov 10, 2017 at 1:37 PM treyd <tr...@treyd.us> wrote:
At some point I plan on figuring out how to do more intelligent front-end
in-browser stuff (Angular, etc) which yeah would definitely help me here but I
was wondering if there was a Django-only way of doing this.
On Friday, November 10, 2017 at 3:51:09 PM UTC-5, Adam wrote:
Are you open to using JavaScript in the front end? It would make it really easy
On Fri, Nov 10, 2017 at 12:40 PM treyd <tr...@treyd.us> wrote:
Hello,
I am trying to render a model field in a django template, and, based on the
value of the field, I want to specify a different CSS class. Right now, my
solution is this:
<span class="label {% if myobject.state == 'Scheduled' %}label-default
{% elif myobject.state == 'Provisioning' %}label-primary
{% elif myobject.state == 'Active' %}label-success
{% elif myobject.state == 'Deleting' %}label-primary
{% elif myobject.state == 'Ended' %}label-info
{% endif %}">{{ myobject.state }}</span>
This works, but seems like a really ugly solution and a bunch of logic in the
template. Is there a better way of doing this? My thought was to put the
class names as another field in the model, but that seems like it would violate
the MVC (or MVT) separation. Any ideas?
Thanks in advance,
treyd
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