Question

我有3张桌子：

class Specialist(models.Model) :
    rating = models.FloatField(default=0.0)

class Service(models.Model) :
    parent_id = models.IntegerField(null=True, blank=True)

class SpecialistPrice(models.Model) :
    specialist = models.ForeignKey(Specialist, related_name='price')
    service = models.ForeignKey(Service, related_name='service')
    price_bottom = models.PositiveSmallIntegerField(null=True, blank=True)

我需要按SpecialistPrice.price_bottom <{1}}以及SpecialistPrice.service.parent_id=0和第2排序条件-rating

的最低对专家进行排序

我需要一个建议。

UPD 我做了：

queryset = queryset.annotate(min_price=Min('price__price_bottom')).order_by('min_price', '-rating')

但得到了错误

Cannot resolve keyword 'min_price' into field.

原始sql工作正常。

 SELECT `specialist`.`id`, `specialist`.`rating`,MIN(`specialistprice`.`price_bottom`) AS `min_price` FROM `specialist` LEFT OUTER JOIN `specialistprice` ON ( `specialist`.`id` = `specialistprice`.`specialist_id` )  GROUP BY `specialist`.`id` ORDER BY `min_price` ASC, `rating` DESC;

Answer 1

您应该阅读lookups that span relationships。此查询应该适合您：

Specialist.objects.filter(price__service__parent_id=0).order_by('price', '-rating')

Django的。正确的分拣方式？

1 个答案: