Graphql 使用石墨烯中继创建订阅

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 =

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 = 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:'