问题
How should one handle a possible race condition in a model's save()
method?
For example, the following example implements a model with an ordered list of related items. When creating a new Item the current list size is used as its position.
From what I can tell, this can go wrong if multiple Items are created concurrently.
class OrderedList(models.Model):
# ....
@property
def item_count(self):
return self.item_set.count()
class Item(models.Model):
# ...
name = models.CharField(max_length=100)
parent = models.ForeignKey(OrderedList)
position = models.IntegerField()
class Meta:
unique_together = (('parent','position'), ('parent', 'name'))
def save(self, *args, **kwargs):
if not self.id:
# use item count as next position number
self.position = parent.item_count
super(Item, self).save(*args, **kwargs)
I've come across @transactions.commit_on_success()
but that seems to apply only to views. Even if it did apply to model methods, I still wouldn't know how to properly handle a failed transaction.
I am currenly handling it like so, but it feels more like a hack than a solution
def save(self, *args, **kwargs):
while not self.id:
try:
self.position = self.parent.item_count
super(Item, self).save(*args, **kwargs)
except IntegrityError:
# chill out, then try again
time.sleep(0.5)
Any suggestions?
Update:
Another problem with the above solution is that the while
loop will never end if IntegrityError
is caused by a name
conflict (or any other unique field for that matter).
For the record, here's what I have so far which seems to do what I need:
def save(self, *args, **kwargs):
# for object update, do the usual save
if self.id:
super(Step, self).save(*args, **kwargs)
return
# for object creation, assign a unique position
while not self.id:
try:
self.position = self.parent.item_count
super(Step, self).save(*args, **kwargs)
except IntegrityError:
try:
rival = self.parent.item_set.get(position=self.position)
except ObjectDoesNotExist: # not a conflict on "position"
raise IntegrityError
else:
sleep(random.uniform(0.5, 1)) # chill out, then try again
回答1:
It may feel like a hack to you, but to me it looks like a legitimate, reasonable implementation of the "optimistic concurrency" approach -- try doing whatever, detect conflicts caused by race conditions, if one occurs, retry a bit later. Some databases systematically uses that instead of locking, and it can lead to much better performance except under systems under a lot of write-load (which are quite rare in real life).
I like it a lot because I see it as a general case of the Hopper Principle: "it's easy to ask forgiveness than permission", which applies widely in programming (especially but not exclusively in Python -- the language Hopper is usually credited for is, after all, Cobol;-).
One improvement I'd recommend is to wait a random amount of time -- avoid a "meta-race condition" where two processes try at the same time, both find conflicts, and both retry again at the same time, leading to "starvation". time.sleep(random.uniform(0.1, 0.6))
or the like should suffice.
A more refined improvement is to lengthen the expected wait if more conflicts are met -- this is what is known as "exponential backoff" in TCP/IP (you wouldn't have to lengthen things exponentially, i.e. by a constant multiplier > 1 each time, of course, but that approach has nice mathematical properties). It's only warranted to limit problems for very write-loaded systems (where multiple conflicts during attempted writes happen quite often) and it may likely not be worth it in your specific case.
回答2:
Add optional FOR UPDATE clause to QuerySets http://code.djangoproject.com/ticket/2705
回答3:
I use Shawn Chin's solution and it proves very useful. The only change I did was to replace the
self.position = self.parent.item_count
with
self.position = self.parent.latest('position').position
just to make sure I am dealing with the latest position number (which in my case might not be item_count because of some reserved unused positions)
来源:https://stackoverflow.com/questions/3522827/handling-race-condition-in-model-save