Question

我一直试图让这个查询工作几个小时，我很乐意帮助，因为我还是Django的新手。

我有这些模型（简化）

class Job(models.Model):
    country = models.ForeignKey(Country)
    active = models.BooleanField()

class Country(models.Model):
    name = models.CharField(max_length=100)

class State(models.Model):
    name = models.CharField(max_length=100)
    country = models.ForeignKey(Country)

class City(models.Model):
    name = models.CharField(max_length=100)
    state = models.ForeignKey(State)

class Worker(models.Model):
    name = models.CharField(max_length=100)
    addresses = models.ManyToManyField(Address)

class Address(models.Model):
    address_field = models.CharField(max_length=100)
    city = models.ForeignKey(City)

我想在我的工人有一个地址的所有国家/地区选择所有不同的和有效的工作。我的观点有这段代码：

def get(self, request):
    worker = Worker.objects.get(pk=request.user.id)
    full_query = Q(active=True)
    # my problem is here, since it is a ManyToMany Relationship
    full_query &= Q(country__id__in=worker.addresses ??????)
    jobs = Job.objects.filter(full_query)

我认为这样可以在for循环中获取所有工作人员的地址，并将国家/地区ID保存在列表中并执行

 country_ids = []
 for address in worker.addresses:
     country_ids.append(address.city.state.country.id)
 Q(country__id__in=[country_ids])

但我认为Django有更优雅的方式。

提前致谢！

Answer 1

尝试

jobs = Job.objects.filter(active=True).filter(country__state__city__address__in=worker.addresses.all())

Django中的复杂内连接

1 个答案: