Question

我在每次比赛中都为同一位足球运动员添加进球而遇到麻烦。

我有课程

class Team(models.Model):
    name = models.CharField()

class Footballer(models.Model):
    name = models.CharField()
    name_team = models.ForeignKey(Team)

class Match(models.Model):
    number = models.IntegerField()
    who = models.ForeingKey(Team,related_name="as_host")
    guest = models.ForeingKey(Team,related_name="as_guest")

class Footballer_stat(models.Model):
    id_footballer = models.ForeignKey(Footballer)
    scored_goals = models.IntegerField('Strzelone Bramki')

我使用来自view.py的视图，它看起来像：

def how_much_goals(request):
    x = Footballer.objects.all()
    z = Footballer_stat.objects.order_by('-scored_goals') 
    sum = Pilkarz_stat.objects.aggregate(Sum("scored_goals"))

    return render_to_response("goals.html",{"footballer":x,"footballer_goals":z, "Shooted_goals":sum["scored_goals__sum"]})

然后在goals.html中，我希望从每个匹配项目中收集目标，例如

Footballer Goals      Results of adding goals the same ballers
    Xxxx     2            Xxxx 3
    Yyyy     3      (+)   Yyyy 4
    Xxxx     1      --->
    Yyyy     1

我应该怎样做才能达到预期的效果？有什么想法吗？

Answer 1

如果你在评论中说have to look like that，那么我可能无法帮助你，你可以忽略这个答案。如果是doesnt have to look like that，那么可能的解决方案就是：

class Team(models.Model):
    name = models.CharField()

class Player(models.Model):
    name = models.CharField()
    team = models.ForeignKey(Team)

class Match(models.Model):
    number = models.IntegerField()
    who = models.ForeingKey(Team,related_name="as_host")
    guest = models.ForeingKey(Team,related_name="as_guest")

class PlayerStats(models.Model):
    player = models.ForeignKey(Player)
    goals = models.IntegerField('Strzelone Bramki')
    match = models.ForeignKey(Match)

    class Meta:
        unique_together = ('player', 'match')

现在，您可以非常轻松地为每场比赛添加每个球员的得分。此外，您只需要将一个QuerySet传递给您的模板：

qs = PlayerStats.objects.select_related().all().extra(Sum('goals'))
qs.group_by = ('player', 'match')
for player in qs:
    qs.player.name # player's name
    qs.goals__sum # goals per match

自从我使用Django以来已经很长时间了，所以它可能不准确。

Django从数据库中收集元素

1 个答案: