如何通过.save()方法将kwargs参数发送到post_save信号

Question

我有两个模型，Process和Notification，我需要在保存流程时创建一个通知，所以我使用postrongave信号：

def create_notification(sender, instance, *args, **kwargs):
if instance.status in ["ERR", "FIN"] or instance.percent == 100:
type = kwargs["type"]
notification = Notification()
notification.account = instance.owner
notification.process = instance
notification.status = "UNREAD"
notification.content = instance.description
notification.notification_type = type
if instance.sub_processes.exists():
for sub_process in instance.sub_processes.all():
if sub_process.model_id != "":
notification.model_id = sub_process.model_id
if sub_process.model_name != "":
notification.model_name = sub_process.model_name
notification.save()

这很好，但我的问题是，当我保存进程时，我需要传递一个额外的参数，这个参数是针对我创建的通知对象，即type，所以每次我保存进程对象(例如在视图中(时，我都试图做类似于process.save(type="EMM")的事情，但它不起作用，create_Notification信号中的行type = kwargs["type"]什么都没做，那么我如何发送额外的参数来保存方法呢？

Answer 1

也许这可以工作：

# Process model
def save(self, type=None, *args, **kwargs):
...

self._type = type

return super(Process, self).save(*args, **kwargs)

# Process post_save
def create_notification(sender, instance, *args, **kwargs):
if instance._type is None:
return
if instance.status in ["ERR", "FIN"] or instance.percent == 100:
type = instance._type
notification = Notification()
notification.account = instance.owner
notification.process = instance
notification.status = "UNREAD"
notification.content = instance.description
notification.notification_type = type
if instance.sub_processes.exists():
for sub_process in instance.sub_processes.all():
if sub_process.model_id != "":
notification.model_id = sub_process.model_id
if sub_process.model_name != "":
notification.model_name = sub_process.model_name
notification.save()