Django信号，在保存列表器上实现 [英] Django signals, implementing on save listner

问题描述

我在理解信号的工作方式时遇到困难，我浏览了一些页面，但没有一个页面帮助我获得图像.

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，我希望孩子在我的应用程序中侦听任何父母，因为这个孩子尤其是通用外键.

core/models.py

django.db导入模型中的

 从django.contrib.auth.models导入用户从django.contrib.contenttypes.models导入ContentType从django.contrib.contenttypes导入通用类Audit(models.Model):## TODO:文档#通过DJANGO内容类型使用通用关系的多态模型操作= 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 =通用.GenericForeignKey('content_type'，'object_id') 

workflow/models.py

django.db导入模型中的

 从django.contrib.auth.models导入用户从django.contrib.contenttypes.models导入ContentType从django.contrib.contenttypes导入通用从core.models导入审核类Instances(models.Model):## TODO:文档## TODO:将ID替换为XXXX-XXXX-XXXX-XXXX# 关于INSTANCE_STATUS =((我"，进行中")，("C"，已取消")，("D"，已删除")，("P"，待处理")，("O"，已完成"))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(审核，editable = False，null = True，空白= True)def save(self，* args，** kwargs):#在新记录上生成一个新的uuid如果self.id为None或self.id .__ len __()为0:导入uuidself.id = uuid.uuid4().__ str __()超级(实例，自我).save(* args，** kwargs)类Setup(models.Model):## TODO:文档#通过DJANGO内容类型使用通用关系的多态模型content_type = models.ForeignKey(ContentType)object_id = models.PositiveIntegerField()content_object = generic.GenericForeignKey('content_type'，'object_id')Actions(models.Model)类:ACTION_TYPE_CHOICES =(("P"，"Python脚本")，("C"，类名")，)名称= models.CharField(max_length = 100)action_type = models.CharField(max_length = 1，choices = ACTION_TYPE_CHOICES)类Tasks(models.Model):名称= models.CharField(max_length = 100)Instance = models.ForeignKey(实例) 

我正在努力在Audit模型中创建一个侦听器，以便我可以将其与Instance模型连接起来.如果在Instance中插入了一条新记录，它也会自动在Audit中插入一个记录.然后，计划将该监听器连接到我的应用中的多个模型，

你知道我该怎么做吗?

解决方案

使用以下代码，您可以使用after_save_instance_handler方法连接实例对象的保存.在这种方法中，您将创建与审核的关系.另请参见通用关系文档

我通常将信号添加到定义了发送方的models.py中.不知道是否需要这样做.

来自django.db.models.signals的

 导入post_save##### SIGNALS ######def after_save_instance_handler(sender，** kwargs):#获取保存的实例instance_object = kwargs ['instance']#获取所需的数据the_date = ...the_user = ...the_object_id = ...#创建与审核对象的关系instance_object.audit_obj.create(operation ="op963"，operation_at = the_date，operation_by = the_user，object_id = the_object_id)#使用后保存信号连接处理程序-Django 1.1 + 1.2post_save.connect(after_save_instance_handler，sender = Instances) 

django 1.2信号

在Django开发版本中，他们添加了一个装饰器来连接信号.因此，除了上面的调用，您必须将此装饰器添加到处理程序中

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

django开发信号

am having difficulties understanding how signals works, I went through some pages but none of them helped me get the picture.

I have two models, I would like to create a signal that will save in the child model when a record is saved in the parent. Actually, I want the child to be listening across my application for any parent since this child in particular of a generic foreign key.

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')

workflow/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
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)

Am struggling to create a listener in Audit model so I can connect it with Instance model, If a new record inserted in Instance it will automatically inserts one in Audit as well. Then, am planning to connect this listener to several models in my app,

Any idea how I can do such a thing?

解决方案

With the code below you can connect the save of an Instance object, with the after_save_instance_handler method. In this method you create the relation to the Audit. Please also see the generic relations doc

I usually add the signals in the models.py where the sender has been defined. Not sure if this is needed.

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)

django 1.2 signals

in django development version they added a decorator to connect the signal. thus instead of the call above you have to add this decorator to the handler

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

django dev signals

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