Reputation: 1614
I have written a python script in my project. I want to update the value of a field.
Here are my modes
class News_Channel(models.Model):
name = models.TextField(blank=False)
info = models.TextField(blank=False)
image = models.FileField()
website = models.TextField()
total_star = models.PositiveIntegerField(default=0)
total_user = models.IntegerField()
class Meta:
ordering = ["-id"]
def __str__(self):
return self.name
class Count(models.Model):
userId = models.ForeignKey(User, on_delete=models.CASCADE)
channelId = models.ForeignKey(News_Channel, on_delete=models.CASCADE)
rate = models.PositiveIntegerField(default=0)
def __str__(self):
return self.channelId.name
class Meta:
ordering = ["-id"]
This is my python script:
from feed.models import Count, News_Channel
def run():
for i in range(1, 11):
news_channel = Count.objects.filter(channelId=i)
total_rate = 0
for rate in news_channel:
total_rate += rate.rate
print(total_rate)
object = News_Channel.objects.filter(id=i)
print(total_rate)
print("before",object[0].total_star,total_rate)
object[0].total_star = total_rate
print("after", object[0].total_star)
object.update()
After counting the total_rate from the Count table I want to update the total star value in News_Channel table. I am failing to do so and get the data before the update and after the update as zero. Although total_rate has value.
Upvotes: 4
Views: 18393
Reputation: 476557
The reason why this fails is because here object
is a QuerySet
of News_Channel
s, yeah that QuerySet
might contain exactly one News_Channel
, but that is irrelevant.
If you then use object[0]
you make a query to the database to fetch the first element and deserialize it into a News_Channel
object. Then you set the total_star
of that object, but you never save that object. You only call .update()
on the entire queryset, resulting in another independent query.
You can fix this with:
objects = News_Channel.objects.filter(id=i)
object = objects[0]
object.total_star = total_rate
object.save()
Or given you do not need any validation, you can boost performance with:
News_Channel.objects.filter(id=i).update(total_star=total_rate)
News_Channel
sIf you want to update all News_Channel
s, you actually better use a Subquery
here:
from django.db.models import OuterRef, Sum, Subquery
subq = Subquery(
Count.objects.filter(
channelId=OuterRef('id')
).annotate(
total_rate=Sum('rate')
).order_by('channelId').values('total_rate')[:1]
)
News_Channel.objects.update(total_star=subq)
Upvotes: 12
Reputation: 6096
The reason is that object
in your case is a queryset, and after you attempt to update object[0]
, you don't store the results in the db, and don't refresh the queryset. To get it to work you should pass the field you want to update into the update method.
So, try this:
def run():
for i in range(1, 11):
news_channel = Count.objects.filter(channelId=i)
total_rate = 0
for rate in news_channel:
total_rate += rate.rate
print(total_rate)
object = News_Channel.objects.filter(id=i)
print(total_rate)
print("before",object[0].total_star,total_rate)
object.update(total_star=total_rate)
print("after", object[0].total_star)
Upvotes: 6
Reputation: 32244
News_Channel.total_star
can be calculated by using aggregation
news_channel_obj.count_set.aggregate(total_star=Sum('rate'))['total_star']
You can then either use this in your script:
object.total_star = object.count_set.aggregate(total_star=Sum('rate'))['total_star']
Or if you do not need to cache this value because performance is not an issue, you can remove the total_star field and add it as a property on the News_Channel
model
@property
def total_star(self):
return self.count_set.aggregate(total_star=Sum('rate'))['total_star']
Upvotes: 2