django暂时禁用信号

Question

我在django中有一个信号回调:

@receiver(post_save, sender=MediumCategory)
def update_category_descendants(sender, **kwargs):

    def children_for(category):
        return MediumCategory.objects.filter(parent=category)

    def do_update_descendants(category):
        children = children_for(category)
        descendants = list() + list(children)

        for descendants_part in [do_update_descendants(child) for child in children]:
            descendants += descendants_part

        category.descendants.clear()
        for descendant in descendants:
            if category and not (descendant in category.descendants.all()):
                category.descendants.add(descendant)
                category.save()
        return list(descendants)

    # call it for update
    do_update_descendants(None)

但是在函数体中我正在使用.save()的模型MediumCategory可以再次发送信号.我该如何禁用它; 完美的解决方案是with内部有一些"魔法" 的声明.

更新: 如果有人感兴趣,这是最终的解决方案.

class MediumCategory(models.Model):
    name = models.CharField(max_length=100)
    slug = models.SlugField(blank=True)
    parent = models.ForeignKey('self', blank=True, null=True)
    parameters = models.ManyToManyField(AdvertisementDescriptonParameter, blank=True)
    count_mediums = models.PositiveIntegerField(default=0)
    count_ads = models.PositiveIntegerField(default=0)

    descendants = models.ManyToManyField('self', blank=True, null=True)

    def save(self, *args, **kwargs):
        self.slug = slugify(self.name)
        super(MediumCategory, self).save(*args, **kwargs)

    def __unicode__(self):
        return unicode(self.name)
(...)
@receiver(post_save, sender=MediumCategory)
def update_category_descendants(sender=None, **kwargs):
    def children_for(category):
        return MediumCategory.objects.filter(parent=category)

    def do_update_descendants(category):
        children = children_for(category)
        descendants = list() + list(children)

        for descendants_part in [do_update_descendants(child) for child in children]:
            descendants += descendants_part

        if category:
            category.descendants.clear()
            for descendant in descendants:
                category.descendants.add(descendant)
        return list(descendants)

    # call it for update
    do_update_descendants(None)

Answer 1

断开信号的@danihp不是DRY和一致的解决方案,例如使用update()而不是save().

要禁用模型上的信号,一个简单的方法是在当前实例上设置属性以防止即将发出的信号.

这可以使用一个简单的装饰器来完成,该装饰器检查给定的实例是否具有'skip_signal'属性,如果是,则阻止调用该方法:

from functools import wraps

def skip_signal():
    def _skip_signal(signal_func):
        @wraps(signal_func)
        def _decorator(sender, instance, **kwargs):
            if hasattr(instance, 'skip_signal'):
                return None
            return signal_func(sender, instance, **kwargs)  
        return _decorator
    return _skip_signal

您现在可以这样使用它:

from django.db.models.signals import post_save
from django.dispatch import receiver

@receiver(post_save, sender=MyModel)
@skip_signal()
def my_model_post_save(sender, instance, **kwargs):
    instance.some_field = my_value
    # Here we flag the instance with 'skip_signal'
    # and my_model_post_save won't be called again
    # thanks to our decorator, avoiding any signal recursion
    instance.skip_signal  = True
    instance.save()

希望这可以帮助.

Answer 2

也许我错了,但我认为category.save()你的代码中不需要这样,add()就足够了,因为改变是在后代但在类别中进行的.

另外,要避免信号,您可以:

• 断开信号并重新连接.
• 使用更新:Descendant.objects.filter( pk = descendant.pk ).update( category = category )