通过builder的Django复杂查询

Question

我已经有一段时间无法使用 ORM 进行复杂的查询了。我知道这是可能的，所以我会原谅帮助。

class Game(models.Model):
    no matter

class Competition(models.Model):
    game = models.ForeignKey(to='game.Game', verbose_name=_('game'), related_name='competitions',
                         on_delete=models.PROTECT)

class User(models.Mode):
    no matter

class Balance(models.Model):
    user = models.OneToOneField(to=User, on_delete=models.CASCADE, primary_key=True)

class BalanceTransaction(models.Model):
    TYPES = (
        (TYPE_COMMISSION, _('commission')),
    }
    competition = models.ForeignKey(to='competition.Competition', verbose_name=_('competition'), related_name='transactions', on_delete=models.CASCADE, null=True, blank=True)
    balance = models.ForeignKey(to=Balance, on_delete=models.CASCADE, related_name='transactions')
    amount = models.DecimalField(_('Transaction amount'), default='0.0', max_digits=28, decimal_places=18,
                             null=False, blank=False)

    type = models.CharField(_('type'), max_length=10, default=TYPE_DEPOSIT, choices=TYPES, null=False, blank=False

我必须 return 一个游戏查询集，其中每个元素都有一个额外的 "income" 字段，用于存储类型为 "commission"[ 的该游戏的所有交易总和=12=]

    Game.objects.filter(publisher__owner=self.request.user.pk).annotate(
    income=Sum(Case(
        When(
            Q(publisher__owner__balance__transactions__type=BalanceTransaction.TYPE_COMMISSION),
            then=???,
        )
    ))

这种方式好像不行，只能用子查询了，目前还没成功

Answer 1

我认为您不需要子查询或大小写，只需一个标准过滤器。

Game.objects.filter(publisher__owner=self.request.user.pk).filter(
    publisher__owner__balance__transactions__type=BalanceTransaction.TYPE_COMMISSION
).annotate(
    income=Sum('publisher__owner__balance__transactions__amount)
)

请参阅 order of annotate and filter clauses 以了解有关其工作原理的解释。

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