如果这在其他地方得到了回答,那么我很抱歉,但下班后 2 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,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()
谜题的最后一部分是事实 事实是附加到 NPC 的一些信息,但是一个人可以知道一个 NPC,但不一定所有与 NPC 相关的事实都被玩家知道,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,它会给我这个:
{
allPlayers {
name
knownNPCs {
image
name
description
factsKnown {
fact
}
}
}
}
其中factsKnown 只是基于Fact 对象的ManyToMany 关系的那些。
到目前为止,我创建的内容会返回数据,但不会根据玩家父级过滤 Facts 仅显示与 npc 相关的所有事实 :(
事实图式
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()
播放器架构:
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)