Question

看看这个模型（它是假设的）：

class Manufacturer(models.Model):
    #...

class Car(models.Model):
    manufacturer = models.ForeignKey(Manufacturer)
    #...

class City(models.Model):
    #...

class Manager(models.Model):
    manufacturer = models.ForeignKey(Manufacturer)
    city = models.ForeignKey(City)
    #...

我想要查询的是：汽车及其制造商的经理列表（给出一些在问题中不重要的条件）以及他们的城市。这可以通过以下代码完成：

manager_car = defaultdict(list)
cars = Car.objects.select_related('manufacturer').filter(...)
for car in cars:
    managers = car.manufacturer.manager_set.select_related('city').filter(...)
    for manager in managers:
        #if <optional condition>:
        manager_car[manager].append(car)

将列在字典中，由经理列出制造商的汽车，这就是我想要的。但是，此代码显然会在数据库中运行与汽车一样多的查询。

如何一次选择所有这些实例？

Answer 1

可能是这样的事情（只是在我头顶编码）：

cars = Car.objects.filter(...)
managers = {}
for manager in Manager.objects.filter(manufacturer__car__in=cars):
    manufacturers = managers.setdefault(manager.manufacturer_id, [])
    manufacturers.append(manager)
cars = list(cars)
for car in cars:
    car.managers = managers.get(car.manufacturer_id, [])

与Django的关系很艰难

1 个答案: