Question

foo_package: ^1.2.3-beta

上面的代码结尾，我在creator_payouts = self.filter(paid=False) processed_user_ids = [] if not creator_payouts.exists(): // here I check the existence return for creator_payout in creator_payouts: do_something() creator_payouts.update(paid=True) // maybe this does something? CreatorRecord.objects.filter( user__in=processed_user_ids, created_for__lte=creator_payouts[0].created_for // error here ).update(processed_for_payout=True)处出现索引错误。

为什么会收到错误消息，尤其是通过上述creator_payouts[0]条件排除了空列表情况

Answer 1

好吧，当您更新creator_payouts时，数据库就会被paid=True更新。由于creator_payouts是一个惰性查询集，当您调用creator_payouts[0]对其求值时，它将获得空查询集。因此，您需要通过强制评估将初始值存储在某个位置。像这样：

if not creator_payouts.exists():   // here I check the existence
         return

for creator_payout in creator_payouts:
    # or store the first creator_payout in a variable here
    do_something()

unpaid_creator_payouts_list = list(creator_payouts)  # force evaluation

creator_payouts.update(paid=True)


 CreatorRecord.objects.filter(
         user__in=processed_user_ids,
         created_for__lte=unpaid_creator_payouts_list[0].created_for // or use the first stored payout_creator in previous loop
     ).update(processed_for_payout=True)

或在CreatorRecord过滤器之后运行更新操作。

Answer 2

creator_payouts是queryset。如果想先变得更好，请尝试使用creator_payouts.first()而不是creator_payouts[0]。

 CreatorRecord.objects.filter(
     user__in=processed_user_ids,
     created_for__lte=creator_payouts.first().created_for
 ).update(processed_for_payout=True)

Django，Queryset存在，但访问[0]会导致索引错误

2 个答案: