如果在其他地方回答了这个问题，那么我很抱歉，但是下班后两天仍然没有雪茄……

我有一个播放器模型：

class Player(models.Model):
    name = models.CharField(max_length=60)
    discord_id = models.CharField(max_length=60,null=True)
    known_npcs = models.ManyToManyField(NPC)

玩家可以认识很多NPC，并且任何NPC都可以被很多玩家认识。

NPC没什么特别的：

class NPC(models.Model):
    image = models.ImageField()
    name = models.CharField(max_length=50)
    description = models.TextField()

难题的最后一部分是事实，事实是与NPC相关的某些信息，但是一个人可以知道NPC，但不一定与玩家有关的事实都与NPC有关，因此事实看起来像这样：

class Fact(models.Model):
    fact = models.TextField()
    known_by = models.ManyToManyField(Player)
    npc = models.ForeignKey(NPC,on_delete=models.DO_NOTHING,null=True)

现在在石墨烯中，我想创建一个Player和allPlayers查询，这将为我提供：

{
  allPlayers {
    name
    knownNPCs {
      image
      name
      description
      factsKnown {
        fact
      }
    }
  }
}

已知事实仅仅是基于Fact对象中的ManyToMany关系的事实。

到目前为止，我创建的内容返回了数据，但未根据玩家父级过滤事实，仅显示了与npc相关的所有事实：(

事实架构

class FactType(DjangoObjectType):
    class Meta:
        model = Fact
        filter_fields = ["id"]

class Query(object):
    fact = Node.Field(FactType)
    all_Facts = graphene.List(FactType)

    def resolve_all_Facts(self,info,**kwargs):
        return Fact.objects.all()

NPCSchema

class NPCType(DjangoObjectType):
    class Meta:
        model = NPCS

class Query(object):
    all_NPCs = graphene.Field(NPCType)
    facts = graphene.List(FactType)
    def resolve_all_NPCs(self,**kwargs):
        return NPCS.objects.all()

PlayerSchema：

class PlayerType(DjangoObjectType):
    class Meta:
        model = Player
        interfaces = (Node,)
        filter_fields = ["id"]


class Query(object):
    player = Node.Field(PlayerType)
    all_players = graphene.List(PlayerType)

    def resolve_all_players(self,**kwargs):
        return Player.objects.all()

    def resolve_player(self,**kwargs):
        player = Player.objects.filter(id=info.id)