类型错误：无法读取未定义和 const gmailEmail = functions.config().gmail.email 的属性“电子邮件”；

问题描述

VisualSTd 从今天起就向我展示了这个错误“TypeError：无法读取未定义的属性‘email’”。

更具体地说，它告诉我“ const gmailEmail = functions.config().gmail.email;” ^ 我是 Jason 的新手，我不太清楚。我在这里写错了，我一次又一次地收到这个错误。不幸的是，我不知道我可以改变什么。

'use strict'

 const functions = require('firebase-functions');
 const nodemailer = require('nodemailer');
 const admin = require('firebase-admin');
admin.initializeApp();




exports.addMessage = functions.https.onRequest(async (req,res) => {

  const original = req.query.text;

  const writeResult = await admin.firestore().collection('messages').add({original: original});

  res.json({result: `Message with ID: ${writeResult.id} added.`});

});

exports.makeUppercase = functions.firestore.document('/messages/{documentId}')
    .onCreate((snap,context) => {

      const original = snap.data().original;

      functions.logger.log('Uppercasing',context.params.documentId,original);
      
      const uppercase = original.toupperCase();
      return snap.ref.set({uppercase},{merge: true});
  
    });






 const gmailEmail = functions.config().gmail.email;
 const gmailPassword = functions.config().gmail.password;

 const mailTransport = nodemailer.createTransport({
    service : 'gmail',auth : {
        user : gmailEmail,pass : gmailPassword,},});



 const APP_NAME = 'InstaClonemm'

 exports.sendWelcomeEmail = functions.auth.user().onCreate((user)  => {           
        const email = user.email;
        const displayName = user.displayName;

    return this.sendWelcomeEmail(email,displayName);
 });


 async function sendWelcomeEmail(email,displayName)  {            
     const mailOptions = {
         from : `${APP_NAME} <[e-mail]`,to : email,};

     mailOptions.subject = `Welcome to ${APP_NAME}!`;               
        mailOptions.text = `hey ${diasplayName || ''}! Welcome to ${APP_NAME},we hope you enjoy our service.`;        


        await mailTransport.sendMail(mailOptions);
        console.log('New email sent to:',email);             


        return null;

       
 }


 

解决方法

暂无找到可以解决该程序问题的有效方法，小编努力寻找整理中！

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