Firebase 3.0.1 Web API查询无法正常工作,出现iFrame安全错误 [英] Firebase 3.0.1 web API querying isn't working, getting an iFrame security error

查看:63
本文介绍了Firebase 3.0.1 Web API查询无法正常工作,出现iFrame安全错误的处理方法,对大家解决问题具有一定的参考价值,需要的朋友们下面随着小编来一起学习吧!

问题描述

我对 ref.once('value',callback)进行了简单的调用.回调从未被调用.

I did a simple call to ref.once('value', callback). The callback never got called.

相反,我在Chrome控制台中收到此错误:拒绝显示'https://console.firebase.google.com/project/project6...redacted...73&parent=http%3A%在框架中2F%2Flocalhost%3A8204& pfname =& rpctoken = 3 ... acteded ... 4',因为它将'X-Frame-Options'设置为'DENY'.

Instead I got this error in my Chrome console: Refused to display 'https://console.firebase.google.com/project/project6...redacted...73&parent=http%3A%2F%2Flocalhost%3A8204&pfname=&rpctoken=3...redacted...4' in a frame because it set 'X-Frame-Options' to 'DENY'.

我不知道这个iFrame是什么,或者这就是为什么我没有得到回调的原因.FWIW,我在 localhost:8204

I have no idea what this iFrame is or whether that's why I didn't get a callback. FWIW, I'm running my dev app on localhost:8204

这是更完整的代码示例.

Here's a fuller code sample.

import firebase from 'firebase/app'
import 'firebase/auth'
import 'firebase/database'

firebase.initializeApp({
  apiKey: config.firebaseApiKey,
  authDomain: `${config.firebaseAppName}.firebaseio.com`,
  databaseURL: `https://${config.firebaseAppName}.firebaseio.com`,
  storageBucket: config.firebaseStorageBucket
})

const ref = firebase.database().ref()

ref.once('value', callback)

推荐答案

好吧,问题出在初始化配置 authDomain 字段中,我写了 .firebaseio.com ,但是它应该是 .firebaseapp.com .

Ok the problem was simply in the initialization config authDomain field, I wrote .firebaseio.com but it should have been .firebaseapp.com.

这篇关于Firebase 3.0.1 Web API查询无法正常工作,出现iFrame安全错误的文章就介绍到这了,希望我们推荐的答案对大家有所帮助,也希望大家多多支持IT屋!

查看全文
登录 关闭
扫码关注1秒登录
发送“验证码”获取 | 15天全站免登陆