Graphql 使用石墨烯中继创建订阅
我不熟悉graphQL石墨烯(python)。我想知道是否可以使用石墨烯创建订阅根类型。谢谢这绝对是可能的。您可以在此处找到有关如何执行此操作的示例: 以下是该回购协议的一个例子:Graphql 使用石墨烯中继创建订阅,graphql,graphene-python,graphql-subscriptions,graphql-mutation,Graphql,Graphene Python,Graphql Subscriptions,Graphql Mutation,我不熟悉graphQL石墨烯(python)。我想知道是否可以使用石墨烯创建订阅根类型。谢谢这绝对是可能的。您可以在此处找到有关如何执行此操作的示例: 以下是该回购协议的一个例子: import asyncio import graphene class Query(graphene.ObjectType): base = graphene.String() class Subscription(graphene.ObjectType): count_seconds =
import asyncio
import graphene
class Query(graphene.ObjectType):
base = graphene.String()
class Subscription(graphene.ObjectType):
count_seconds = graphene.Float(up_to=graphene.Int())
async def resolve_count_seconds(root, info, up_to):
for i in range(up_to):
yield i
await asyncio.sleep(1.)
yield up_to
schema = graphene.Schema(query=Query, subscription=Subscription)
这绝对是可能的。您可以在此处找到有关如何执行此操作的示例: 以下是该回购协议的一个例子:
import asyncio
import graphene
class Query(graphene.ObjectType):
base = graphene.String()
class Subscription(graphene.ObjectType):
count_seconds = graphene.Float(up_to=graphene.Int())
async def resolve_count_seconds(root, info, up_to):
for i in range(up_to):
yield i
await asyncio.sleep(1.)
yield up_to
schema = graphene.Schema(query=Query, subscription=Subscription)
这在执行期间给了我以下错误:
订阅必须返回Async Iterable或Observable。已接收:“
在执行过程中出现以下错误:订阅必须返回Async Iterable或Observable。收到:'