如何在django中更新queryset值？

Question

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.

Answer 1

The problem

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.total_star = total_rate
object

Or given you do not need any validation, you can boost performance with:

News_Channel.objects.filter(id=i).update(total_star=total_rate)

Updating all News_Channel s

If 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)

Answer 2

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)

Answer 3

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']