I have a model:
from django.db import models
CHOICES = (
('s', 'Glorious spam'),
('e', 'Fabulous eggs'),
)
class MealOrder(models.Model):
meal = models.CharField(max_length=8, choices=CHOICES)
I have a form:
from django.forms import ModelForm
class MealOrderForm(ModelForm):
class Meta:
model = MealOrder
And I want to use formtools.preview. The default template prints the short version of the choice ('e' instead of 'Fabulous eggs'), becuase it uses
{% for field in form %}
<tr>
<th>{{ field.label }}:</th>
<td>{{ field.data }}</td>
</tr>
{% endfor %}.
I'd like a template as general as the mentioned, but printing 'Fabulous eggs' instead.
[as I had doubts where's the real question, I bolded it for all of us :)]
I know how to get the verbose version of a choice in a way that is itself ugly:
{{ form.meal.field.choices.1.1 }}
The real pain is I need to get the selected choice, and the only way coming to my mind is iterating through choices and checking {% ifequals currentChoice.0 choiceField.data %}
, which is even uglier.
Can it be done easily? Or it needs some template-tag programming? Shouldn't that be available in django already?
In Django templates you can use the "get_FOO_display()
" method, that will return the readable alias for the field, where 'FOO' is the name of the field.
Note: in case the standard FormPreview
templates are not using it, then you can always provide your own templates for that form, which will contain something like {{ form.get_meal_display }}
.
The best solution for your problem is to use helper functions. If the choices are stored in the variable CHOICES and the model field storing the selected choice is 'choices' then you can directly use
{{ x.get_choices_display }}
in your template. Here, x is the model instance. Hope it helps.
My apologies if this answer is redundant with any listed above, but it appears this one hasn't been offered yet, and it seems fairly clean. Here's how I've solved this:
from django.db import models
class Scoop(models.Model):
FLAVOR_CHOICES = [
('c', 'Chocolate'),
('v', 'Vanilla'),
]
flavor = models.CharField(choices=FLAVOR_CHOICES)
def flavor_verbose(self):
return dict(Scoop.FLAVOR_CHOCIES)[self.flavor]
My view passes a Scoop to the template (note: not Scoop.values()), and the template contains:
{{ scoop.flavor_verbose }}
Basing on Noah's reply, here's a version immune to fields without choices:
#annoyances/templatetags/data_verbose.py
from django import template
register = template.Library()
@register.filter
def data_verbose(boundField):
"""
Returns field's data or it's verbose version
for a field with choices defined.
Usage::
{% load data_verbose %}
{{form.some_field|data_verbose}}
"""
data = boundField.data
field = boundField.field
return hasattr(field, 'choices') and dict(field.choices).get(data,'') or data
I'm not sure wether it's ok to use a filter for such purpose. If anybody has a better solution, I'll be glad to see it :) Thank you Noah!
We can extend the filter solution by Noah to be more universal in dealing with data and field types:
<table>
{% for item in query %}
<tr>
{% for field in fields %}
<td>{{item|human_readable:field}}</td>
{% endfor %}
</tr>
{% endfor %}
</table>
Here's the code:
#app_name/templatetags/custom_tags.py
def human_readable(value, arg):
if hasattr(value, 'get_' + str(arg) + '_display'):
return getattr(value, 'get_%s_display' % arg)()
elif hasattr(value, str(arg)):
if callable(getattr(value, str(arg))):
return getattr(value, arg)()
else:
return getattr(value, arg)
else:
try:
return value[arg]
except KeyError:
return settings.TEMPLATE_STRING_IF_INVALID
register.filter('human_readable', human_readable)
I don't think there's any built-in way to do that. A filter might do the trick, though:
@register.filter(name='display')
def display_value(bf):
"""Returns the display value of a BoundField"""
return dict(bf.field.choices).get(bf.data, '')
Then you can do:
{% for field in form %}
<tr>
<th>{{ field.label }}:</th>
<td>{{ field.data|display }}</td>
</tr>
{% endfor %}
You have Model.get_FOO_display() where FOO is the name of the field that has choices.
In your template do this :
{{ scoop.get_flavor_display }}
Add to your models.py one simple function:
def get_display(key, list):
d = dict(list)
if key in d:
return d[key]
return None
Now, you can get verbose value of choice fields like that:
class MealOrder(models.Model):
meal = models.CharField(max_length=8, choices=CHOICES)
def meal_verbose(self):
return get_display(self.meal, CHOICES)
Upd.: I'm not sure, is that solution “pythonic” and “django-way” enough or not, but it works. :)
The extended-extended version of Noah's and Ivan's solution. Also fixed Noah's solution for Django 3.1, as ModelChoiceIteratorValue
is now unhashable.
@register.filter
def display_value(value: Any, arg: str = None) -> str:
"""Returns the display value of a BoundField or other form fields"""
if not arg: # attempt to auto-parse
# Returning regular field's value
if not hasattr(value.field, 'choices'): return value.value()
# Display select value for BoundField / Multiselect field
# This is used to get_..._display() for a read-only form-field
# which is not rendered as Input, but instead as text
return list(value.field.choices)[value.value()][1]
# usage: {{ field|display_value:<arg> }}
if hasattr(value, 'get_' + str(arg) + '_display'):
return getattr(value, 'get_%s_display' % arg)()
elif hasattr(value, str(arg)):
if callable(getattr(value, str(arg))):
return getattr(value, arg)()
return getattr(value, arg)
return value.get(arg) or ''
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