Python Django信号，在save listner上实现

我很难理解信号是如何工作的，我浏览了一些页面，但没有一页能帮助我理解

我有两个模型，我想创建一个信号，当记录保存在父模型中时，该信号将保存在子模型中。实际上，我希望该子对象在我的应用程序中侦听任何父对象，因为该子对象特别是通用外键

core/models.py

from django.db import models
from django.contrib.auth.models import User
from django.contrib.contenttypes.models import ContentType
from django.contrib.contenttypes import generic

class Audit(models.Model):
    ## TODO: Document
    # Polymorphic model using generic relation through DJANGO content type
    operation  = models.CharField(max_length=40)
    operation_at = models.DateTimeField("Operation At", auto_now_add=True)
    operation_by = models.ForeignKey(User, db_column="operation_by", related_name="%(app_label)s_%(class)s_y+")
    content_type = models.ForeignKey(ContentType)
    object_id = models.PositiveIntegerField()
    content_object = generic.GenericForeignKey('content_type', 'object_id')

from django.db import models
from django.contrib.auth.models import User
from django.contrib.contenttypes.models import ContentType
from django.contrib.contenttypes import generic
from core.models import Audit


class Instances(models.Model):
    ##  TODO: Document
    ##  TODO: Replace id with XXXX-XXXX-XXXX-XXXX
    # Re
    INSTANCE_STATUS = (
        ('I', 'In Progress' ),
        ('C', 'Cancelled'   ),
        ('D', 'Deleted'     ),
        ('P', 'Pending'     ),
        ('O', 'Completed'   )
    )

    id=models.CharField(max_length=200, primary_key=True)
    status=models.CharField(max_length=1, choices=INSTANCE_STATUS, db_index=True)
    audit_obj=generic.GenericRelation(Audit, editable=False, null=True, blank=True)


    def save(self, *args, **kwargs):
        # on new records generate a new uuid
        if self.id is None or self.id.__len__() is 0:
            import uuid
            self.id=uuid.uuid4().__str__()
        super(Instances, self).save(*args, **kwargs)



class Setup(models.Model):
    ## TODO: Document
    # Polymorphic model using generic relation through DJANGO content type
    content_type=models.ForeignKey(ContentType)
    object_id=models.PositiveIntegerField()
    content_object=generic.GenericForeignKey('content_type', 'object_id')


class Actions(models.Model):
    ACTION_TYPE_CHOICES = (
        ('P', 'Python Script'),
        ('C', 'Class name'),
    )
    name=models.CharField(max_length=100)
    action_type=models.CharField(max_length=1, choices=ACTION_TYPE_CHOICES)


class Tasks(models.Model):
    name=models.CharField(max_length=100)
    Instance=models.ForeignKey(Instances)

工作流/模型.py

from django.db import models
from django.contrib.auth.models import User
from django.contrib.contenttypes.models import ContentType
from django.contrib.contenttypes import generic

class Audit(models.Model):
    ## TODO: Document
    # Polymorphic model using generic relation through DJANGO content type
    operation  = models.CharField(max_length=40)
    operation_at = models.DateTimeField("Operation At", auto_now_add=True)
    operation_by = models.ForeignKey(User, db_column="operation_by", related_name="%(app_label)s_%(class)s_y+")
    content_type = models.ForeignKey(ContentType)
    object_id = models.PositiveIntegerField()
    content_object = generic.GenericForeignKey('content_type', 'object_id')

from django.db import models
from django.contrib.auth.models import User
from django.contrib.contenttypes.models import ContentType
from django.contrib.contenttypes import generic
from core.models import Audit


class Instances(models.Model):
    ##  TODO: Document
    ##  TODO: Replace id with XXXX-XXXX-XXXX-XXXX
    # Re
    INSTANCE_STATUS = (
        ('I', 'In Progress' ),
        ('C', 'Cancelled'   ),
        ('D', 'Deleted'     ),
        ('P', 'Pending'     ),
        ('O', 'Completed'   )
    )

    id=models.CharField(max_length=200, primary_key=True)
    status=models.CharField(max_length=1, choices=INSTANCE_STATUS, db_index=True)
    audit_obj=generic.GenericRelation(Audit, editable=False, null=True, blank=True)


    def save(self, *args, **kwargs):
        # on new records generate a new uuid
        if self.id is None or self.id.__len__() is 0:
            import uuid
            self.id=uuid.uuid4().__str__()
        super(Instances, self).save(*args, **kwargs)



class Setup(models.Model):
    ## TODO: Document
    # Polymorphic model using generic relation through DJANGO content type
    content_type=models.ForeignKey(ContentType)
    object_id=models.PositiveIntegerField()
    content_object=generic.GenericForeignKey('content_type', 'object_id')


class Actions(models.Model):
    ACTION_TYPE_CHOICES = (
        ('P', 'Python Script'),
        ('C', 'Class name'),
    )
    name=models.CharField(max_length=100)
    action_type=models.CharField(max_length=1, choices=ACTION_TYPE_CHOICES)


class Tasks(models.Model):
    name=models.CharField(max_length=100)
    Instance=models.ForeignKey(Instances)

我正在努力在审计模型中创建一个侦听器，这样我就可以将它与实例模型连接起来，如果在实例中插入一条新记录，它也会自动在审计中插入一条。然后，我计划将此侦听器连接到我应用程序中的几个模型

---

你知道我该怎么做吗？

通过下面的代码，你可以用after\u save\u Instance\u handler方法连接实例对象的保存。在此方法中，您可以创建与审核的关系。另请参阅

我通常在models.py中添加信号，其中定义了发送方。不确定是否需要这样做

from django.db.models.signals import post_save

#####SIGNALS######
def after_save_instance_handler(sender, **kwargs):
    #get the saved instance
    instance_object = kwargs['instance']

    #get the needed data
    the_date = ...
    the_user = ...
    the_object_id = ...

    #create the relation to the audit object
    instance_object.audit_obj.create(operation="op963",operation_at=the_date,operation_by=the_user,object_id=the_object_id)

#connect the handler with the post save signal - django 1.1 + 1.2
post_save.connect(after_save_instance_handler, sender=Instances)

在django开发版本中，他们添加了一个装饰器来连接信号。 因此，您必须将此装饰程序添加到处理程序中，而不是上面的调用

@receiver(post_save, sender=Instances)
def after_save_instance_handler(sender, **kwargs):

---

使用下面的代码，您可以使用after_save_Instance_handler方法连接实例对象的保存。在此方法中，您可以创建与审核的关系。另请参阅

我通常在models.py中添加信号，其中定义了发送方。不确定是否需要这样做

from django.db.models.signals import post_save

#####SIGNALS######
def after_save_instance_handler(sender, **kwargs):
    #get the saved instance
    instance_object = kwargs['instance']

    #get the needed data
    the_date = ...
    the_user = ...
    the_object_id = ...

    #create the relation to the audit object
    instance_object.audit_obj.create(operation="op963",operation_at=the_date,operation_by=the_user,object_id=the_object_id)

#connect the handler with the post save signal - django 1.1 + 1.2
post_save.connect(after_save_instance_handler, sender=Instances)

在django开发版本中，他们添加了一个装饰器来连接信号。 因此，您必须将此装饰程序添加到处理程序中，而不是上面的调用

@receiver(post_save, sender=Instances)
def after_save_instance_handler(sender, **kwargs):

---

重要的是：为什么？由于子级对父级具有FK，因此子级的save可以简单地调用父级的save。你不需要信号。重要的是：为什么？由于子级对父级具有FK，因此子级的save可以简单地调用父级的save。你不需要信号。汤姆，你写的代码太棒了！我已经用一些型号试过了，效果非常好。尽管如此，我在使用具体的模型实例时遇到了一个问题。显然，因为我正在实例内部导入/使用Audit类，所以当我尝试在Audit内部导入实例时，Django会抛出一个错误。它无法看到我得到的类导入错误如下：无法导入名称实例不客气！那么您想在Audit中导入实例？这些模型是否放置在相同的models.py中？如果是，审计后是否定义了实例模型？如果是的话，我会想到一个想法。不确定是否导入，但至少通过设置ForeignKey，引用的模型必须放在引用模型之上。Tom，你编写的代码太棒了！我已经用一些型号试过了，效果非常好。尽管如此，我在使用具体的模型实例时遇到了一个问题。显然，因为我正在实例内部导入/使用Audit类，所以当我尝试在Audit内部导入实例时，Django会抛出一个错误。它无法看到我得到的类导入错误如下：无法导入名称实例不客气！那么您想在Audit中导入实例？这些模型是否放置在相同的models.py中？如果是，审计后是否定义了实例模型？如果是的话，我会想到一个想法。不确定是否导入，但至少通过设置ForeignKey，引用的模型必须放置在引用模型的上方。