一个相当复杂的 Django 查询

Question

我有

class Supplier(Model) : 
   pass

class Customer(Model) : 
   pass

class Dock(Model) : 
   pass

class SupplierDockAccess(Model) : 
   supplier = ForeignKey(Supplier)
   dock = ForeignKey(Dock)

class SupplierCustomerAccess(Model):
   supplier = ForeignKey(Supplier)
   customer = ForeignKey(Customer)

我有一个客户实例，我想获取客户有权访问的所有 Docks。客户可以通过 SupplierCustomerAccess 访问供应商，供应商可以通过 SupplierDockAccess 访问 Docks。我可以这样做：

# get the suppliers the customer has access to
supplier_customer_accesses = SupplierCustomerAccess.objects.filter(customer=customer)
suppliers = [s.supplier for s in supplier_customer_accesses]

# get the docks those suppliers have access to
supplier_dock_accesses = SupplierDockAccess.objects.filter(supplier__in=suppliers)
docks = [s.dock for s in supplier_dock_accesses]

...但是结果的码头列表包含重复项，我真的认为应该可以一口气完成。有人想展示一些强大的 django-fu 吗？

Answer 1

我能想到的最简单的方法是结合ManyToManyFields 和custom QuerySet/Manager。

from django.db import models
class CustomQuerySetManager(models.Manager):
    """
        Class for making QuerySet methods available on result set
        or through the objects manager.
    """
    def get_query_set(self):   
        return self.model.QuerySet(self.model) 
    def __getattr__(self, attr, *args): 
        try:                   
             return getattr(self.__class__, attr, *args)
        except AttributeError:
             return getattr(self.get_query_set(), attr, *args)

class Customer(models.Model):
   suppliers = models.ManyToManyField(through=SupplierCustomerAccess)
   objects = CustomQuerySetManager()
   class QuerySet(QuerySet):
       def docks(self):
           return Dock.objects.filter(
                   supplierdockaccess__supplier__in=self.suppliers
               ).distinct()
   ...
class Supplier(models.Model):
   docks = models.ManyToManyField(through=SupplierDockAccess)
   ...

您应该只看到一两次对数据库的点击（取决于您在获得客户时是否使用了select_related），并且您的代码非常干净：

docks = customer.docks()
suppliers = customer.suppliers.all()
...

Answer 2

好吧，我想通了。谈论它似乎可以解决问题的事情之一：

docks = Dock.objects.filter(supplierdockaccess__supplier__suppliercustomeraccess__customer=customer).distinct()

...看看 sql，它确实是在一个大连接中完成的。好的。抱歉回答我自己的问题。