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Compose choices keys using ContentType name not text label

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This commit is contained in:
Roberto Rosario
2012-10-05 00:51:00 -04:00
parent cd05f7fde0
commit e69750dd49
1 changed files with 4 additions and 4 deletions
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8
apps/common/utils.py
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@@ -315,16 +315,16 @@ def parse_range(astr):
def generate_choices_w_labels(choices, display_object_type=True):
results = []
for choice in choices:
ct_label = ContentType.objects.get_for_model(choice).name
ct = ContentType.objects.get_for_model(choice)
label = unicode(choice)
if isinstance(choice, User):
label = choice.get_full_name() if choice.get_full_name() else choice
if display_object_type:
verbose_name = unicode(getattr(choice._meta, u'verbose_name', ct_label))
results.append((u'%s,%s' % (ct_label, choice.pk), u'%s: %s' % (verbose_name, label)))
verbose_name = unicode(getattr(choice._meta, u'verbose_name', ct.name))
results.append((u'%s,%s' % (ct.model, choice.pk), u'%s: %s' % (verbose_name, label)))
else:
results.append((u'%s,%s' % (ct_label, choice.pk), u'%s' % (label)))
results.append((u'%s,%s' % (ct.model, choice.pk), u'%s' % (label)))
#Sort results by the label not the key value
return sorted(results, key=lambda x: x[1])