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.
来源:https://stackoverflow.com/questions/11487128/django-temporarily-disable-signals