【发布时间】:2018-09-11 06:39:25
【问题描述】:
我正在创建一个推送通知应用程序,我使用 node.js 来部署 firebase 功能,但在部署时显示错误。
警告避免嵌套承诺承诺/不嵌套
警告避免嵌套承诺承诺/不嵌套
error 每个 then() 都应该返回或抛出 promise/always-return
这是我的代码:
"use-strict";
const functions = require("firebase-functions");
const admin = require("firebase-admin");
admin.initializeApp(functions.config().firebase);
exports.sendNotification = functions.firestore
.document("Users/{user_id}/Notification/{notification_id}")
.onWrite(event => {
const user_id = event.params.user_id;
const notification_id = event.params.notification_id;
return admin
.firestore()
.collection("Users")
.doc(user_id)
.collection("Notification")
.doc(notification_id)
.get()
.then(queryResult => {
const from_user_id = queryResult.data().from;
const from_message = queryResult.data().message;
const from_data = admin
.firestore()
.collection("Users")
.doc(from_user_id)
.get();
const to_data = admin
.firestore()
.collection("Users")
.doc(user_id)
.get();
return Promise.all([from_data, to_data]).then(result => {
const from_name = result[0].data().name;
const to_name = result[1].data().name;
const token_id = result[1].data().token_id;
const payload = {
notification: {
title: "Notification From :" + from_name,
body: from_message,
icon: "default"
}
};
return admin
.messaging()
.sendToDevice(token_id, payload)
.then(result => {
console.log("Notification Sent.");
});
});
});
});
【问题讨论】:
-
这里函数的格式和缩进让人难以阅读。
标签: javascript node.js firebase google-cloud-functions eslint