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我在 Google App Engine 柔性环境中运行 firebase。我在一个节点上设置了一个子添加侦听器，每次添加一个子时它都会被调用两次。这只发生在我在 Google App Engine 上运行此代码时。如果我在本地运行它，它会按预期运行。这是代码

// [START app]
'use strict';

var express = require('express');
var bodyParser = require('body-parser');
var request = require('request');
var firebase = require('firebase-admin');

var app = express();

firebase.initializeApp({
    credential: firebase.credential.applicationDefault(),
    databaseURL: "https://myurl.firebaseio.com/"
});

var db = firebase.database();
var globalNotifications = db.ref("/globalNotifications");

var API_KEY = "myKey"; // Your Firebase Cloud Server API key
var now = new Date();

app.use(bodyParser.json());
app.use(bodyParser.urlencoded({ extended: false }));

globalNotifications.orderByChild('processed').equalTo(0).on("child_added", function(snapshot) {

    var key = snapshot.key;
    var incomingNotification = snapshot.val();
    var userID = incomingNotification.userID;
    var notifyID = incomingNotification.notifyID;

    request({
        url: 'https://fcm.googleapis.com/fcm/send',
        method: 'POST',
        headers: {
            'Content-Type' :'application/json',
            'Authorization': 'key='+API_KEY
        },
        body: JSON.stringify({
            "to" : incomingNotification.notificationID,
            "priority" : "high",
            "notification" : {
                "body" : "someone is following you",
                "title" : "You have a new follower"
            }
        })
    }, function(error, response, body) {
        if (error) { console.error(error); }
        else if (response.statusCode >= 400) {
            console.error('HTTP Error: '+response.statusCode+' - '+response.statusMessage);
        }
        else {
            console.log(response.statusCode);
            console.log(body);
            globalNotifications.child(key).update({"processed": 1, "response": body});
        }
    });

});

// Start the server
var server = app.listen(process.env.PORT || '8080', function () {
    console.log('App listening on port %s', server.address().port);
    console.log('Press Ctrl+C to quit.');
});
// [END app]

感谢您提前提供任何帮助。

4

1 回答 1

7

经过一段时间的测试，我明白了。此问题是由在 Google App Engine 上运行 Firebase-Admin 引起的。本质上，您的 AppEngine 所在的每个底层虚拟机都在运行自己的 Firebase-Admin 实例。默认情况下，AppEngine 为每个服务至少维护两个 VM。因此，在测试中（在最小负载下），两个 VM 实例中的每一个都会调用一次 firebase 函数。

显然，没有关于这个问题的文档，但是 Google/Firebase 有一个库可以解决这个问题。它被称为 Firebase-Queue，可以在 NPM 上找到。我用这段代码解决了我的问题：

// [START notificationsservice app]
'use strict';

var express = require('express');
var bodyParser = require('body-parser');
var request = require('request');
var rp = require('request-promise');
var admin = require('firebase-admin');
var Queue = require('firebase-queue');

var app = express();

admin.initializeApp({
    credential: admin.credential.cert("serviceAccountCredentials.json"),
    databaseURL: "https://<YOUR PROJECT ID HERE>.firebaseio.com/"
});

var db = admin.database();
var notifications = db.ref('/notifications');

var API_KEY = "<YOUR API KEY HERE>"

app.use(bodyParser.json());
app.use(bodyParser.urlencoded({ extended: false }));

var queue = new Queue(notifications, function(data, progress, resolve, reject) {

    var incomingNotification = data;
    var userID = incomingNotification.userID;
    var username = incomingNotification.username;

    rp({
        url: 'https://fcm.googleapis.com/fcm/send',
        method: 'POST',
        headers: {
            'Content-Type' :'application/json',
            'Authorization': 'key='+API_KEY
        },
        body: JSON.stringify({
            "to" : "/topics/follower-"+userID,
            "priority": "high",
            "notification" : {
                "body" : username+" you have a notification",
            },
            "data" : {
                "type" : "follower"
            }
        })
    }).then(function(body) {
        progress(100);
        console.log("Notification sent."+body);
        resolve();
    }).catch(function(error) {
        progress(21);
        console.log(error);
        reject(); 
    });

  setTimeout(function() {
    resolve();
  }, 1000);
});

// Start the server
var server = app.listen(process.env.PORT || '8080', function () {
    console.log('App listening on port %s', server.address().port);
    console.log('Press Ctrl+C to quit.');
});
// [END app]
于 2016-12-05T21:47:51.077 回答