Question

I want to update field:number with 'current number + random number' :

Here is my code, I have 650 games in database,and it needs almost 8 seconds to update it.

objs = Game.objects.all()
for obj in objs:
    Game.objects.filter(name=obj.name).update(number=F('number') + random.randint(1,100))

If I use F function, the the code below will have a problem :
the random.randint(1,100) is the same for all rows ,
Can I randomly use different numbers with F funciton? Or Is there some method by which I can speed up my query ??

Game.objects.all().update(number=F('number') + random.randint(1,100))

Answer 1

是的，您可以先在第一步生成随机数，然后将它们分配给Game模型的数字字段，像这样：

objs = Game.objects.all()
for obj in objs:
    random_number = random.randint(1,100)
    obj.number = random_number
    obj.save()

或者以更好的方式，你不需要遍历for循环，只需使用它：

def generate_random(request):
  number = random.randint(1,100)
  return number

def your_view(request):

  objs = Game.objects.all()
  objs.update(number=F('number') + generate_random() )

感谢。

Can Django F function bulk update random number?

1 个答案: