Question

我正在编写一个联盟系统，我希望在每个赛季中显示每个球员在该赛季累积的积分排名。

到目前为止，我设法使用与此类似的代码：

class Player(models.Model):
    name = models.CharField(max_length=100)


class Season(models.Model):
    name = models.CharField(max_length=100)
    start_date = models.DateField()
    end_date = models.DateField()
    players = models.ManyToManyField(Player)

    def get_player_rank(self, player):
        return player.matchresult_set.filter(season=self).aggregate(points=Sum('points'))['points']

    def get_ranks(self):
        ranks = [(player, self.get_player_rank(player)) for player in self.players.all()]
        ranks.sort(key=lambda tup: tup[1])
        return ranks


class Match(models.Model):
    date = models.DateField()
    players = models.ManyToManyField(Player, through='MatchResult')
    season = models.ForeignKey(Season)


class MatchResult(models.Model):
    player = models.ForeignKey(Player)
    match = models.ForeignKey(Match)
    points = models.IntegerField()

我认为通过更简单的聚合可以实现同样的目标，但我无法正确获得annotate（）。

我试过这个但是它只是总结了整个赛季的所有要点：

class Season(models.Model):    
    def get_ranks(self):
        return self.players.annotate(points=Sum('matchresult__points')).order_by('-points')

我错过了什么？我猜。如果它会导致可移植的代码，可以使用.extra（）。

Answer 1

这会返回可用的结果：

Season.objects.values('name','match__matchresult__player__username').annotate(points=Sum('match__matchresult__points')).distinct()

我还需要实现SumWithDefault来摆脱NULL：

from django.db.models.sql.aggregates import Aggregate
from django.db.models import Aggregate as Ag

class SumWithDefaultSQL(Aggregate):
    def __init__(self, col, default=None, **extra):
        super(SumWithDefaultSQL, self).__init__(col, default=default, **extra)
        self.sql_function = 'SUM'
        if default is not None:
            self.sql_template = 'COALESCE(%(function)s(%(field)s), %(default)s)'


class SumWithDefault(Ag):
    name = 'Sum'

    def add_to_query(self, query, alias, col, source, is_summary):
        aggregate = SumWithDefaultSQL(col, source=source, is_summary=is_summary, **self.extra)
        query.aggregates[alias] = aggregate

最终查询：

Season.objects.values('name','match__matchresult__player__username').annotate(points=SumWithDefault('match__matchresult__points', default=0)).distinct().order_by('-points')

Answer 2

Django的ORM并未涵盖所有用例。

您可以将聚合存储在单独的模型中，也可以回退到原始SQL。

Django通过QuerySet聚合？

2 个答案: