- Author:
- sterkh
- Posted:
- July 10, 2012
- Language:
- Python
- Version:
- 1.3
- Tags:
- checkboxselectmultiple widget disabled
- Score:
- 0 (after 0 ratings)
Inspired by http://djangosnippets.org/snippets/2453/
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 | from itertools import chain
from django.forms.widgets import Select, CheckboxSelectMultiple, CheckboxInput, mark_safe
from django.utils.encoding import force_unicode
from django.utils.html import escape, conditional_escape
class CheckboxSelectMultipleWithDisabled(CheckboxSelectMultiple):
"""
Subclass of Django's checkbox select multiple widget that allows disabling checkbox-options.
To disable an option, pass a dict instead of a string for its label,
of the form: {'label': 'option label', 'disabled': True}
"""
def render(self, name, value, attrs=None, choices=()):
if value is None: value = []
has_id = attrs and 'id' in attrs
final_attrs = self.build_attrs(attrs, name=name)
output = [u'<ul>']
# Normalize to strings
str_values = set([force_unicode(v) for v in value])
for i, (option_value, option_label) in enumerate(chain(self.choices, choices)):
if final_attrs.has_key('disabled'):
del final_attrs['disabled']
if isinstance(option_label, dict):
if dict.get(option_label, 'disabled'):
final_attrs = dict(final_attrs, disabled = 'disabled' )
option_label = option_label['label']
# If an ID attribute was given, add a numeric index as a suffix,
# so that the checkboxes don't all have the same ID attribute.
if has_id:
final_attrs = dict(final_attrs, id='%s_%s' % (attrs['id'], i))
label_for = u' for="%s"' % final_attrs['id']
else:
label_for = ''
cb = CheckboxInput(final_attrs, check_test=lambda value: value in str_values)
option_value = force_unicode(option_value)
rendered_cb = cb.render(name, option_value)
option_label = conditional_escape(force_unicode(option_label))
output.append(u'<li><label%s>%s %s</label></li>' % (label_for, rendered_cb, option_label))
output.append(u'</ul>')
return mark_safe(u'\n'.join(output))
|
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