我想为Firebase创建多个云功能,并从一个项目同时部署它们。我还想将每个函数分离到一个单独的文件中。目前,我可以创建多个函数,如果我把它们都放在index.js,如:
exports.foo = functions.database.ref('/foo').onWrite(event => {
...
});
exports.bar = functions.database.ref('/bar').onWrite(event => {
...
});
然而,我想把foo和酒吧在单独的文件。我试了一下:
/functions
|--index.js (blank)
|--foo.js
|--bar.js
|--package.json
foo.js在哪里
exports.foo = functions.database.ref('/foo').onWrite(event => {
...
});
bar.js是
exports.bar = functions.database.ref('/bar').onWrite(event => {
...
});
有没有一种方法可以在不把所有函数都放在index.js中的情况下实现这一点?
Firebase文档现在已经更新了一个多文件代码组织的好指南:
文档>云功能>写功能>组织功能
总结:
foo.js
const functions = require('firebase-functions');
exports.foo = functions.https.onRequest((request, response) => {
// ...
});
bar.js
const functions = require('firebase-functions');
exports.bar = functions.https.onRequest((request, response) => {
// ...
});
index.js
const foo = require('./foo');
const bar = require('./bar');
exports.foo = foo.foo;
exports.bar = bar.bar;
啊,Firebase负载节点模块的云函数通常,所以这是有效的
结构:
/functions
|--index.js
|--foo.js
|--bar.js
|--package.json
index.js:
const functions = require('firebase-functions');
const fooModule = require('./foo');
const barModule = require('./bar');
exports.foo = functions.database.ref('/foo').onWrite(fooModule.handler);
exports.bar = functions.database.ref('/bar').onWrite(barModule.handler);
foo.js:
exports.handler = (event) => {
...
};
bar.js:
exports.handler = (event) => {
...
};
更新:Typescript现在完全支持,所以不需要下面的恶作剧。只需使用firebase cli
以下是我个人是如何使用typescript的:
/functions
|--src
|--index.ts
|--http-functions.ts
|--main.js
|--db.ts
|--package.json
|--tsconfig.json
在此之前,我先提出两点警告:
进口/出口的顺序在index.ts中很重要
db必须是一个单独的文件
第二点,我不知道为什么。其次,你应该完全尊重我的index, main和db的配置(至少要尝试一下)。
索引。Ts:与出口有关。我觉得让索引更干净。贸易部门负责出口。
// main must be before functions
export * from './main';
export * from "./http-functions";
主要。ts:处理初始化。
import { config } from 'firebase-functions';
import { initializeApp } from 'firebase-admin';
initializeApp(config().firebase);
export * from "firebase-functions";
db。Ts:只是重新导出数据库,所以它的名字比database()短。
import { database } from "firebase-admin";
export const db = database();
http-functions.ts
// db must be imported like this
import { db } from './db';
// you can now import everything from index.
import { https } from './index';
// or (both work)
// import { https } from 'firebase-functions';
export let newComment = https.onRequest(createComment);
export async function createComment(req: any, res: any){
db.ref('comments').push(req.body.comment);
res.send(req.body.comment);
}
这种格式允许您的入口点查找其他函数文件,并自动导出每个文件中的每个函数。
主要入口脚本
找到functions文件夹中的所有.js文件,并导出从每个文件中导出的每个函数。
const fs = require('fs');
const path = require('path');
// Folder where all your individual Cloud Functions files are located.
const FUNCTIONS_FOLDER = './scFunctions';
fs.readdirSync(path.resolve(__dirname, FUNCTIONS_FOLDER)).forEach(file => { // list files in the folder.
if(file.endsWith('.js')) {
const fileBaseName = file.slice(0, -3); // Remove the '.js' extension
const thisFunction = require(`${FUNCTIONS_FOLDER}/${fileBaseName}`);
for(var i in thisFunction) {
exports[i] = thisFunction[i];
}
}
});
从一个文件导出多个函数
Const functions = require('firebase-functions');
Const query = functions.https。onRequest((req, res) => {
Let query = req.query.q;
res.send ({
“You searching For”:查询
});
});
const searchTest = functions.https。onRequest((req, res) => {
res.send ({
"searchTest": "你好!"
});
});
模块。出口= {
查询
searchTest
}
HTTP可访问端点有适当的命名
✔功能:查询:http://localhost:5001/PROJECT-NAME/us-central1/query
function: helloWorlds: http://localhost:5001/PROJECT-NAME/us-central1/helloWorlds
✔功能:searchTest: http://localhost:5001/PROJECT-NAME/us-central1/searchTest
一个文件
如果你只有几个额外的文件(例如只有一个),你可以使用:
Const your_functions = require('./path_to_your_functions');
For (var I in your_functions) {
export [i] = your_functions[i];
}
@jasonsirota的回答很有帮助。但是查看更详细的代码可能会有用,特别是在HTTP触发函数的情况下。
使用与@jasonsirota回答中相同的结构,假设你希望在两个不同的文件中有两个单独的HTTP触发函数:
目录结构:
/functions
|--index.js
|--foo.js
|--bar.js
|--package.json
index.js:
'use strict';
const fooFunction = require('./foo');
const barFunction = require('./bar');
// Note do below initialization tasks in index.js and
// NOT in child functions:
const functions = require('firebase-functions');
const admin = require('firebase-admin');
admin.initializeApp(functions.config().firebase);
const database = admin.database();
// Pass database to child functions so they have access to it
exports.fooFunction = functions.https.onRequest((req, res) => {
fooFunction.handler(req, res, database);
});
exports.barFunction = functions.https.onRequest((req, res) => {
barFunction.handler(req, res, database);
});
foo.js:
exports.handler = function(req, res, database) {
// Use database to declare databaseRefs:
usersRef = database.ref('users');
...
res.send('foo ran successfully');
}
bar.js:
exports.handler = function(req, res, database) {
// Use database to declare databaseRefs:
usersRef = database.ref('users');
...
res.send('bar ran successfully');
}