Question

我有以下资源，我试图覆盖obj_create。如果我不覆盖它，事情就会完美，但是当我覆盖它时会出现POST错误。任何线索？真的很感激答案虽然我相信tastypie真的没有得到任何堆栈溢出的答案这些天..这是非常烦人的。我正在考虑可能因为相同的原因而切换我的堆栈。

代码如下：

class OrderResource(BackBoneCompatibleResource):
  person = fields.ToOneField(PersonResource, 'person', full=True)
  restaurant = fields.ToOneField(RestaurantResource, 'restaurant', full=True)
  itemList = fields.ToManyField(OrderItemResource, 'itemList', full=True)

  class Meta:
    object_class = Order
    queryset = Order.objects.all().order_by("-time_updated")
    resource_name = 'order'
    allowed_methods = ['get','post','put','delete','patch']
    authorization = Authorization()
    serializer = Serializer(formats=['json', 'jsonp', 'xml', 'yaml', 'html', 'plist'])
    authentication = ClientAuthentication()
    authorization = OrderAuthorization()
    always_return_data = True
    filtering = {
        "restaurant" : ["exact"],
        "time_created" : ["gte"],
        "person" : ["exact"]
    }

  def obj_create(self, bundle, request=None, **kwargs):
    print "Entered Order Create"
    return super(OrderResource, self).obj_create(bundle, request, **kwargs)

订单模型是：

class Order(models.Model):
  restaurant = models.ForeignKey(Restaurant)
  person = models.ForeignKey(Person)
  tableNumber = models.CharField(max_length=2)
  PLACED = 'p'
  ACCEPTED = 'a'
  READY = 'r'
  ORDER_STATUS_CHOICES = (
    (PLACED, 'Placed'),
    (ACCEPTED, 'Accepted'),
    (READY, 'Ready'),
  ) 
  order_status = models.CharField(max_length=1, choices=ORDER_STATUS_CHOICES, default=PLACED)
  itemList = models.ManyToManyField(OrderItem, null=True)
  tax = models.FloatField()
  tip = models.FloatField()
  cost = models.FloatField()
  time_created = models.DateTimeField(auto_now_add=True)
  time_updated = models.DateTimeField(auto_now=True)

正如我所说的，如果从资源中删除obj_create（）函数，则发布正确。我只是编写文档中指定的默认函数，我无法理解我在做什么？我可能会遗漏一些非常明显的东西。谢谢你的时间..

我得到的具体错误如下，不确定它是否有多大帮助..

<type 'exceptions.TypeError'>, TypeError('obj_create() takes exactly 2 arguments (3 given)',), <traceback object at 0x10d30fb90>

Answer 1

更改此行：

return super(OrderResource, self).obj_create(bundle, request, **kwargs)

对此：

return super(OrderResource, self).obj_create(bundle, request=request, **kwargs)

request must be passed as a keyword argument。

Answer 2

当您致电super().obj_create时，您正在向父方法传递一个它并不期望的参数。错误消息告诉您错误，当您查看documentation中的父方法时，它会有意义。它会抓住**kwargs，但看看有没有*args？这意味着它无法处理您传递的request参数。

Resource.obj_create(self, bundle, **kwargs)  # Tastypie

我建议您使用相同的方法签名在您的子类上定义obj_create，这样您就不会混淆调用者的期望。父母没有明确处理request所以你也不应该这样做。最终代码：

class OrderResource(BackBoneCompatibleResource):
    def obj_create(self, bundle, **kwargs):
        print "Entered Order Create"
        return super(OrderResource, self).obj_create(bundle, **kwargs)

Tastypie重写obj_create

2 个答案: