Question

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, info, **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, info, **kwargs):
        return Player.objects.all()

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

Answer 1

对于仍然阅读的人，我设法通过创建“ filteredFacts”字段来解决此问题：

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

filtered_facts = graphene.List(FactGroup)

def resolve_filtered_facts(self, info, **kwargs):
    groups = defaultdict(list)
    facts = self.known_facts.all()
    for fact in facts:
        groups[fact.npc].append(fact.fact)
    grouped_facts = []
    for key, value in groups.items():
        grouped_facts.append(FactGroup(npc=key, facts=value))

    return grouped_facts

这将获得玩家已知的所有事实，并按NPC将其分组，而从另一端得到相同的结果...

Python Graphene具有多对多关系

1 个答案: