I'd like to update multiple integer fields at once in following model.
class Foo(models.Model):
field_a = models.PositiveIntegerField()
field_b = models.PositiveIntegerField()
field_c = models.PositiveIntegerField()
Originally, it can be done like following code with two queries.
foo = Foo.objects.get(id=1)
foo.field_a += 1
foo.field_b -= 1
foo.field_c += 2
foo.save()
I'd like make it more simpler with update in one query.
However, following attempts raised error.
# 1st attempt
Foo.objects.filter(id=1).update(
field_a=F('field_a')+1,
field_b=F('field_a')-1,
field_c=F('field_a')+2)
# 2nd attempt
Foo.objects.filter(id=1).\
update(field_a=F('field_a')+1).\
update(field_b=F('field_b')-1) ).\
update(field_c=F('field_c')+2)
How can I solve this ?
Form the django docs:
Calls to update can also use F expressions to update one field based on the value of another field in the model. This is especially useful for incrementing counters based upon their current value. For example, to increment the pingback count for every entry in the blog:
>>> from django.db.models import F
>>> Entry.objects.all().update(n_pingbacks=F('n_pingbacks') + 1)
You have to have an instance of Foo or a queryset before you can update. You should do something like this:
Foo.objects.get(id=1)update(field_a=F('field_a')+1).\
update(field_b=F('field_b')-1) ).\
update(field_c=F('field_c')+2)
or
Foo.objects.filter(id__in=[1,3,6,7]).update(field_a=F('field_a')+1).\
update(field_b=F('field_b')-1) ).\
update(field_c=F('field_c')+2)
Reference: https://docs.djangoproject.com/en/1.8/topics/db/queries/#updating-multiple-objects-at-once
If save() is passed a list of field names in keyword argument update_fields, only the fields named in that list will be updated. This may be desirable if you want to update just one or a few fields on an object. There will be a slight performance benefit from preventing all of the model fields from being updated in the database. For example:
product.name = 'Name changed again'
product.save(update_fields=['name'])
see more docs [here]:https://docs.djangoproject.com/en/dev/ref/models/instances/#specifying-which-fields-to-save
Related
class Parent(models.Model):
# some fields
class Child(models.Model):
parent = models.ForeginKey(Parent)
species = models.ForeignKey(Species)
# other fields
I have a function like this:
1. def some_function(unique_id):
2. parent_object = Parent.objects.get(unique_id=unique_id)
3. new_child = Child.objects.create(name='Joe', parent=parent_object)
4. call_some_func(new_child.parent.name, new_child.species.name)
In the line 4, a db query is generated for Species. Is there any way, I can use select_related to prefetch the Species, so as to prevent extra query.
Can it be done while I use .create(). This is just an example , I am using many other fields too and they are querying the DB every time.
The only way I can think is after line 3 using this code:
child_obj = Child.objects.select_related('species').get(id=new_child.id)
The only parameter that create accepts is force_insert which is not related to what you're asking, so it seems it's not possible. Also, noticing that create performs an INSERT ... RETURNING ... statement, I don't think it would be possible anyway because you cannot return columns from other tables.
Possibly the best approach is what you already suggested: do a get() afterwards with the related fields you need.
How can I check whether a many-to-many relationship exists or another property is fulfilled? When I try the query, it returns some rows twice!
Given a model
from django.db import models
class Plug(models.Model):
name = models.CharField(primary_key=True, max_length=99)
class Widget(models.Model):
name = models.CharField(primary_key=True, max_length=99)
shiny = models.BooleanField()
compatible = models.ManyToManyField(Plug)
I have the following items in my database:
from django.db.models import Q
schuko = Plug.objects.create(name='F')
uk = Plug.objects.create(name='G')
Widget.objects.create(name='microwave', shiny=True).compatible.set([uk])
Widget.objects.create(name='oven', shiny=False).compatible.set([uk])
Widget.objects.create(name='pc', shiny=True).compatible.set([uk, schuko])
Now I want all names of widgets that are shiny and/or compatible with Schuko:
shiny_or_schuko = sorted(
Widget.objects.filter(Q(shiny=True) | Q(compatible=schuko))
.values_list('name', flat=True))
But to my surprise, this does not return ['microwave', 'pc']. Instead, 'pc' is listed twice, i.e. shiny_or_schuko is ['microwave', 'pc', 'pc'].
Is this a Django bug? If not, how can I set up the query that I get 'pc' just once?
Is this a Django bug?
No. You simply perform a LEFT OUTER JOIN with the many-to-many table. If two or more related objects match, it will be included multiple times. This can be wanted behavior, for example if you add extra annotations to the elements that takes values from these related objects.
You can make use of .distinct() [Django-doc] to return only distinct elements:
Widget.objects.filter(
Q(shiny=True) | Q(compatible=schuko)
).values_list('name', flat=True).distinct()
I have two models:
Model A is an AbstractUserModel and Model B
class ModelB:
user = ForeignKey(User, related_name='modelsb')
timestamp = DateTimeField(auto_now_add=True)
What I want to find is how many users have at least one ModelB object created at least in 3 of the 7 past days.
So far, I have found a way to do it but I know for sure there is a better one and that is why I am posting this question.
I basically split the query into 2 parts.
Part1:
I added a foo method inside the User Model that checks if a user meets the above conditions
def foo(self):
past_limit = starting_date - timedelta(days=7)
return self.modelsb.filter(timestamp__gte=past_limit).order_by('timestamp__day').distinct('timestamp__day').count() > 2
Part 2:
In the Custom User Manager, I find the users that have more than 2 modelsb objects in the last 7 days and iterate through them applying the foo method for each one of them.
By doing this I narrow down the iterations of the required for loop. (basically its a filter function but you get the point)
def boo(self):
past_limit = timezone.now() - timedelta(days=7)
candidates = super().get_queryset().annotate(rc=Count('modelsb', filter=Q(modelsb__timestamp__gte=past_limit))).filter(rc__gt=2)
return list(filter(lambda x: x.foo(), candidates))
However, I want to know if there is a more efficient way to do this, that is without the for loop.
You can use conditional annotation.
I haven't been able to test this query, but something like this should work:
from django.db.models import Q, Count
past_limit = starting_date - timedelta(days=7)
users = User.objects.annotate(
modelsb_in_last_seven_days=Count('modelsb__timestap__day',
filter=Q(modelsb__timestamp__gte=past_limit),
distinct=True))
.filter(modelsb_in_last_seven_days__gte = 3)
EDIT:
This solution did not work, because the distinct option does specify what field makes an entry distinct.
I did some experimenting on my own Django instance, and found a way to make this work using SubQuery. The way this works is that we generate a subquery where we make the distinction ourself.
counted_modelb = ModelB.objects
.filter(user=OuterRef('pk'), timestamp__gte=past_limit)
.values('timestamp__day')
.distinct()
.annotate(count=Count('timestamp__day'))
.values('count')
query = User.objects
.annotate(modelsb_in_last_seven_days=Subquery(counted_modelb, output_field=IntegerField()))
.filter(modelsb_in_last_seven_days__gt = 2)
This annotates each row in the queryset with the count of all distinct days in modelb for the user, with a date greater than the selected day.
In the subquery I use values('timestamp__day') to make sure I can do distinct() (Because a combination of distinct('timestamp__day') and annotate() is unsupported.)
I have these models:
def Foo(Models.model):
size = models.IntegerField()
# other fields
def is_active(self):
if check_condition:
return True
else:
return False
def Bar(Models.model):
foo = models.ForeignKey("Foo")
# other fields
Now I want to query Bars that are having active Foo's as such:
Bar.objects.filter(foo.is_active())
I am getting error such as
SyntaxError at /
('non-keyword arg after keyword arg'
How can I achieve this?
You cannot query against model methods or properties. Either use the criteria within it in the query, or filter in Python using a list comprehension or genex.
You could also use a custom manager. Then you could run something like this:
Bar.objects.foo_active()
And all you have to do is:
class BarManager(models.Manager):
def foo_active(self):
# use your method to filter results
return you_custom_queryset
Check out the docs.
I had similar problem: I am using class-based view object_list and I had to filter by model's method. (storing the information in database wasn't an option because the property was based on time and I would have to create a cronjob and/or... no way)
My answer is ineffective and I don't know how it's gonna scale on larger data; but, it works:
q = Model.objects.filter(...)...
# here is the trick
q_ids = [o.id for o in q if o.method()]
q = q.filter(id__in=q_ids)
You can't filter on methods, however if the is_active method on Foo checks an attribute on Foo, you can use the double-underscore syntax like Bar.objects.filter(foo__is_active_attribute=True)
assume I have this little model:
class Deal(models.Model):
purchases = models.IntegerField(default=0)#amount of purchases so far
increase_purchases(self,to_add):
self.update( purchases =self.purchases + to_add)
when I try to use this increase_purchases model from shell:
>>> x = Deal.objects.get(id=1)
>>> x.increase_purchases(4)
AttributeError: 'Deal' object has no attribute 'update'
How can I write a proper function to the model so that I can update the selected querys purchases as I want ?
Based on your example and description, you probably want something like this:
class Deal(models.Model):
purchase_count = models.IntegerField(default=0)
def purchase(self, quantity=1):
self.purchase_count = self.purchase_count + quantity
I agree with Ignacio; modify the object and then save it. So in the shell:
> great_deal = Deal.objects.get(id=1)
> great_deal.purchase(4)
> great_deal.save()
> # or w/o an explicite argument it will record a single purchase
> # great_deal.purchase()
Yes, I renamed things a little bit in the Deal model. It just seemed more descriptive this way.
Modify the appropriate fields then call save() on the instance.
Or use the += expression for cleaner code:
class Deal(models.Model):
purchase_count = models.IntegerField(default=0)
def purchase(self, quantity=1):
self.purchase_count += quantity
In Django 1.6.2. Encountered this behavior and used a "filter" then update works as expected. For example, Students.objects.select_for_update().filter(id=3).update(score = 10)
Just fyi: Unless you are handling transactions, modifying each field separately using save() might create data inconsistency in a multi-threaded environment. By the time threadA calls save() on a model, another threadB could have changed the model fields and saved. In which case threadA has to read the updated model and change.