【发布时间】:2019-01-12 13:53:17
【问题描述】:
代码
'use strict'
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(event =>{
const user_id = event.params.user_id;
const notification_id = event.params.notification_id;
console.log('We have a notification to send to ', user_id);
if(!event.data.val()){
return console.log("A Notification has been deleted from the database", notification_id);
}
const deviceToken = admin.database().ref(`/UserData/${user_id}/TokenID`).once('value');
return deviceToken.then(result =>{
const 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 =>{
return console.log('This was the notofication Feature');
});
});
});
错误
应用程序的所有代码都可以在 android 中使用 java 完成,但函数应该在 javascript 中完成,这太糟糕了...语言新手,所以不确定错误甚至意味着什么...有人可以帮我解决它请问?
【问题讨论】:
-
如果您是 JavaScript 新手,Cloud Functions for Firebase 并不是学习它的最佳方式。我建议先阅读Firebase documentation for Web developers 和/或阅读Firebase codelab for Web developer。它们涵盖了许多基本的 JavaScript、Web 和 Firebase 交互。您还可以在本地 Node.js 进程中使用 Admin SDK,该进程可以使用本地调试器进行调试。完成这些之后,您将能够更好地为 Cloud Functions 编写代码。
标签: javascript android firebase firebase-realtime-database google-cloud-functions