django temporarily disable signals

问题:

I have a signal callback in django:

@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)

but in the function body I'm using .save() on models MediumCategory that couses that the signal is dispatched again. How can I disable it; the perfect solution would be a with statement with some 'magic' inside.

UPDATE: That's final solution, if anyone interested.

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)

回答1:

Perhaps I'm wrong, but I think that category.save() is not needed in your code, add() is enough because change is made in descendant but in category.

Also, to avoid signals you can:

• Disconnect signal and reconnect.
• Use update: Descendant.objects.filter( pk = descendant.pk ).update( category = category )

回答2:

@danihp disconnecting a signal is not a DRY and consistent solution, such as using update() instead of save().

To disable a signal on your model, a simple way to go is to set an attribute on the current instance to prevent upcoming signals firing.

This can be done using a simple decorator that checks if the given instance has the 'skip_signal' attribute, and if so prevents the method from being called:

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

You can now use it this way:

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()

Hope This helps.