mirror of
https://github.com/jazzband/django-admin-sortable.git
synced 2026-03-17 06:20:33 +00:00
37f91cce97
Refactored how the sortable_by properties get populated by looping over the model fields until we get to the SortableForeignKey, then grabbing properties from the field and its related data.
71 lines
2.4 KiB
Python
71 lines
2.4 KiB
Python
from django.contrib.contenttypes.models import ContentType
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from django.db import models
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from adminsortable.fields import SortableForeignKey
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class MultipleSortableForeignKeyException(Exception):
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def __init__(self, value):
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self.value = value
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def __str__(self):
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return repr(self.value)
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class Sortable(models.Model):
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"""
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Unfortunately, Django doesn't support using more than one AutoField in a model
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or this class could be simplified.
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`is_sortable` determines whether or not the Model is sortable by determining
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if the last value of `order` is greater than the default of 1, which should be
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present if there is only one object.
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`model_type_id` returns the ContentType.id for the Model that inherits Sortable
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`save` the override of save increments the last/highest value of order by 1
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Override `sortable_by` method to make your model be sortable by a foreign key field.
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Set `sortable_by` to the class specified in the foreign key relationship.
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"""
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order = models.PositiveIntegerField(editable=False, default=1, db_index=True)
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#legacy support
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sortable_by = None
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class Meta:
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abstract = True
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ordering = ['order']
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@classmethod
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def is_sortable(cls):
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try:
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max_order = cls.objects.aggregate(models.Max('order'))['order__max']
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except TypeError, IndexError:
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max_order = 0
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return True if max_order > 1 else False
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@classmethod
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def model_type_id(cls):
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return ContentType.objects.get_for_model(cls).id
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def __init__(self, *args, **kwargs):
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super(Sortable, self).__init__(*args, **kwargs)
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#Validate that model only contains at most one SortableForeignKey
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sortable_foreign_keys = []
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for field in self._meta.fields:
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if isinstance(field, SortableForeignKey):
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sortable_foreign_keys.append(field)
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if len(sortable_foreign_keys) > 1:
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raise MultipleSortableForeignKeyException(u'%s may only have one SortableForeignKey' % self)
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def save(self, *args, **kwargs):
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if not self.id:
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try:
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self.order = self.__class__.objects.aggregate(models.Max('order'))['order__max'] + 1
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except TypeError, IndexError:
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pass
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super(Sortable, self).save(*args, **kwargs)
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