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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>>>>>> --
>>>>>> --
>>>>>>
>>>>>> Adam F. Simon, PhD
>>>>>> Calabasas, CA.
>>>>>>
>>>>>> cell: 818-425-3719
>>>>>> home: 818-880-8405
>>>>>>
>>>>>> Feel free to link w/ me: LinkedIn
>>>>>>
>>>>>>
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>>>>>
>>>> --
>>>> --
>>>>
>>>> Adam F. Simon, PhD
>>>> Calabasas, CA.
>>>>
>>>> cell: 818-425-3719
>>>> home: 818-880-8405
>>>>
>>>> Feel free to link w/ me: LinkedIn
>>>>
>>>>
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>>>>
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>>>
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