Firebase云功能的Javascript代码，提供针对推送通知编写的错误

Question

Firebase云功能的Javascript代码，提供针对推送通知编写的错误

index.js

'use strict'
//Install functions and admin sdks'
const functions = require('firebase-functions');
const admin =require('firebase-admin');

admin.initializeApp(functions.config().firebase);

exports.sendNotification = functions.database.ref('/notifications/{user_id}/{notification_id}').onWrite((change, context) => {
    //onwrite will run if data changed
    const user_id = context.params.user_id;
    const notification = context.params.notification;

    if (context.params === null) {
        return console.log("Notification Has Been Deleted");
    }
    let token_id = null;
    const deviceToken = admin.database.ref(`/Users/${user_id}/device_token`).once('value');
    return deviceToken.then(result => {

        token_id = result.val();

        const payload = {
            notification:{
                title:"Friend Request",
                body:"You have recieved a new friend request",
                icon:"default"
            }
        };


    });


    return admin.messaging().sendToDevice(token_id, payload).then(response =>{
        console.log('This is the notify feature');
    }).catch(err => {
        console.log('Error getting documents', err);
    });

});

运行firebase deploy命令时，会出现如下错误。如果有人能支持，我会很感激。

错误

错误每个then（）应返回一个值或抛出promise / always-return 错误无法访问的代码无法访问 error每个then（）应返回一个值或抛出promise / always-return

npm ERR! code ELIFECYCLE
npm ERR! errno 1
npm ERR! functions@ lint: `eslint .`
npm ERR! Exit status 1
npm ERR! 
npm ERR! Failed at the functions@ lint script.
npm ERR! This is probably not a problem with npm. There is likely additional logging output above.

Answer 1

我希望它能正常工作，你应该像这样回归

return console.log('This is the notify feature');

Answer 2

错误是说你应该从传递给任何then（）方法的回调中返回一个值。您的代码中有两个then（）方法。如果您没有任何内容可以发送到promise链中的下一个工作项，只需在函数末尾return null。

但是，您的代码中存在更大的问题。您在函数的顶层有两个return语句，一个在另一个之后。这几乎肯定会导致另一个错误，或者至少不能达到预期的效果。（您必须只返回一个在所有工作完成后解决的承诺。）

Answer 3

道格感谢你的评论。然后我改变了代码。现在它部署没有任何错误。但是在调用方法之后它在firebase函数控制台中给出了一个错误。（调用方法意味着使用android应用程序添加新通知）在有效负载中，没有令牌：deviceToken部分也没有工作

'use strict'


//Install functions and admin sdks'
const functions = require('firebase-functions');
const admin =require('firebase-admin');
var FCM = require('fcm-push');
admin.initializeApp();


var serverKey = 'MY_SERVER_KEY_HERE';
var fcm = new FCM(serverKey);



exports.sendNotification = functions.database.ref('/notifications/{user_id}/{notification_id}').onWrite((change, context) => {
    //onwrite will run if data changed
    const user_id = context.params.user_id;
    const notification = context.params.notification;
console.log('user : '+user_id)
console.log('notification : '+notification)
    // If exit the function.
    if (!change.after.val()) {
      return console.log('User ', user_id, 'notification', notification);
    }

// Only edit data when it is first created.
      if (change.before.exists()) {
        return null;
      }
      // Exit when the data is deleted.
      if (!change.after.exists()) {
        return null;
      }

    const deviceToken = admin.database().ref(`/Users/${user_id}/device_token`).once('value');
    console.log("Dev TOken : "+deviceToken);
    return admin.database().ref(`/Users/${user_id}/device_token`).once('value').then(result => {


        const payload = {
            notification:{
                title:"Friend Request",
                body:"You have recieved a new friend request",
            },
            tokens:deviceToken
        };

        return admin.messaging().send(payload).then(response =>{
            return console.log('This is the notify feature');
        }).catch(err => {
            return console.log('Error getting documents', err);
        });
    })

});

输出： - https://i.imgur.com/9PRwMbi.png

如果有任何错误，任何人都可以更正代码吗？

Answer 4

全部谢谢。我解决了这个问题。

'use strict'

//Install functions and admin sdks'
const functions = require('firebase-functions');
const admin =require('firebase-admin');
admin.initializeApp();

exports.sendNotification = functions.database.ref('/notifications/{user_id}/{notification_id}').onWrite((change, context) => {
    //onwrite will run if data changed
    const user_id = context.params.user_id;
    //const from = context.params.from;
    const notification_id = context.params.notification_id;
    console.log('user : '+user_id);
    console.log('notification_id : '+notification_id);
    //console.log('notification : '+notification)
    //console.log('from : '+from)
    // If exit the function.
    if (!change.after.val()) {
      return console.log('Sender ', user_id, 'From', from);
    }

// Only edit data when it is first created.
      if (change.before.exists()) {
        return null;
      }
      // Exit when the data is deleted.
      if (!change.after.exists()) {
        return null;
      }

    // if (context.params === null) {
    //     return console.log("Notification Has Been Deleted");
    // }

//    let token_id = null;
    const getDeviceTokensPromise = admin.database().ref(`/Users/${user_id}/device_token`).once('value');

    return getDeviceTokensPromise.then(result => {
      const payload = {
        notification: {
          title: 'You have a new follower!',
          body: `is now following you.`,
          icon: 'default'
        }
      };

      return admin.messaging().sendToDevice(result.val(), payload);

    }).then(res => {
      return console.log(JSON.stringify(res));
    });  

});

Answer 5

对于云功能，如果其https触发器，则它必须向用户返回响应，如果还有任何其他类型的触发器，则它必须返回Promise。只是return Promise.resolve("Done");我曾尝试返回null，但这对我不起作用。

Firebase云功能说无法访问，每个then（）都应返回一个值或抛出promise / always-return

Firebase云功能的Javascript代码，提供针对推送通知编写的错误

index.js

错误

5 个答案: