订阅字段必须返回异步Iterable。收到:未确定

Question

我在阿波罗服务器上使用graphql nodejs。当我尝试时，我得到了这个错误：

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我已经在解析器中返回pubsub.asyncIterator了。它会返回错误。另一个项目相同的设置工作，但在那里，它不起作用。我在这里检查了一些堆栈问题，但我无法解决。

服务器：App.js

const express = require('express');
const mongoose = require('mongoose');
const { ApolloServer } = require('apollo-server');


const { mainSchema } = require('./graphql/schemas/index');
const { rootResolver } = require('./graphql/resolvers/index');

const app = express();


const server = new ApolloServer({
typeDefs: mainSchema,
resolvers: rootResolver,
subscriptions: {
         path: `/graphql`,
         onConnect: (connectionParams, webSocket, context) => {
         console.log('Client connected');
},
onDisconnect: (webSocket, context) => {
         console.log('Client disconnected')
    },

  },
});

server.listen(3000, "192.168.1.106").then(({ url }) => {
console.log(`  Server ready at ${url}`);
});

订阅解析器

    Subscription :{
    userNotifications : withFilter(
            ()=>pubSub.asyncIterator(NEW_NOTIFICATION_CREATE),
            (payload, variables)=>{
                console.log("payload=>", payload);
                console.log("variables=>", variables);
                return true
            }
        )}}

PubSub.js

  const { PubSub } = require('graphql-subscriptions')

  const pubSub =new PubSub()

  exports.pubSub = pubSub;

Answer 1

我必须使用subccribe关键字来定义subs。谢谢我。订阅:{ userNotifications : withFilter(子代码:userNotifications(有效载荷，变量)=>{ console.log("payload=>"，有效载荷)；console.log("variables=>"，变量)；返回true }}