Bel*_*dez 9 python django-templates django-models
我的(简化)模型:
class Stop(models.Model):
EXPRESS_STOP = 0
LOCAL_STOP = 1
STOP_TYPES = (
(EXPRESS_STOP, 'Express stop'),
(LOCAL_STOP, 'Local stop'),
)
name = models.CharField(max_length=32)
type = models.PositiveSmallIntegerField(choices=STOP_TYPES)
price = models.DecimalField(max_digits=5, decimal_places=2, null=True, blank=True)
def _get_cost(self):
if self.price == 0:
return 0
elif self.type == self.EXPRESS_STOP:
return self.price / 2
elif self.type == self.LOCAL_STOP:
return self.price * 2
else:
return self.price
cost = property(_get_cost)
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我的目标:我想按照cost物业排序.我尝试了两种方法.
Stops.objects.order_by('cost')
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这产生了以下模板错误:
Caught FieldError while rendering: Cannot resolve keyword 'cost' into field.
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{% with deal_items|dictsort:"cost_estimate" as items_sorted_by_price %}
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收到以下模板错误:
Caught VariableDoesNotExist while rendering: Failed lookup for key [cost] in u'Union Square'
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我应该怎么做呢?
Ign*_*ams 15
使用QuerySet.extra()连同CASE ... END定义一个新的领域,和排序上.
Stops.objects.extra(select={'cost': 'CASE WHEN price=0 THEN 0 '
'WHEN type=:EXPRESS_STOP THEN price/2 WHEN type=:LOCAL_STOP THEN price*2'},
order_by=['cost'])
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那,或者将QuerySet其余的返回到一个列表,然后使用L.sort(key=operator.attrgetter('cost'))它.