按模型的属性(而不是字段)对 Django QuerySet 进行排序 [英] Sorting a Django QuerySet by a property (not a field) of the Model

问题描述

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)

我的目标:我想按 cost 属性排序.我尝试了两种方法.

My Goal: I want to sort by the cost property. I tried two approaches.

Stops.objects.order_by('cost')

这产生了以下模板错误:

That yielded the following template error:

Caught FieldError while rendering: Cannot resolve keyword 'cost' into field.

使用 dictsort 模板过滤器

{% with deal_items|dictsort:"cost_estimate" as items_sorted_by_price %}

收到以下模板错误:

Caught VariableDoesNotExist while rendering: Failed lookup for key [cost] in u'Union Square'

所以...

我应该怎么做?

推荐答案

使用 QuerySet.extra() 与 CASE ... END 一起定义一个新字段，并对其进行排序.

Use QuerySet.extra() along with CASE ... END to define a new field, and sort on that.

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'])

那个，或者将从其余部分返回的 QuerySet 转换为列表，然后在其上使用 L.sort(key=operator.attrgetter('cost')).

That, or cast the QuerySet returned from the rest to a list, then use L.sort(key=operator.attrgetter('cost')) on it.

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