如果这个问题在其他地方得到回答,那么我很抱歉,但是下类后两天,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,而任何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,但是一个人可以认识一个 NPC,但玩家不一定知道所有与 NPC 相关的事实,因此事实是看起来像这样:
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 Graphite 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
}
}
}
}
已知的事实只是基于 Fact 对象中的多对多关系的事实。
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事实,只是显示与 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()
NPC架构
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()
玩家架构:
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)
最佳答案
对于任何仍在阅读的人,我设法通过创建“filteredFacts”字段来解决这个问题:
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
这会获取玩家已知的所有事实并按 NPC 对其进行分组,与另一方的结果相同...
关于Python Graphene 处理多对多关系,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/58718721/