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?


当前回答

我有这个项目,它有后台函数和http函数。我还有用于单元测试的测试。CI/CD将使您在部署云功能时更加轻松

文件夹结构

|-- package.json
|-- cloudbuild.yaml
|-- functions
    |-- index.js
    |-- background
    |   |-- onCreate
    |       |-- index.js
            |-- create.js
    |
    |-- http
    |   |-- stripe
    |       |-- index.js
    |       |-- payment.js
    |-- utils
        |-- firebaseHelpers.js
    |-- test
        |-- ...
    |-- package.json

注意:utils/文件夹用于在函数之间共享代码

函数/ index.js

在这里,您可以导入所需的所有函数并声明它们。这里不需要逻辑。在我看来,这样更干净。

require('module-alias/register');
const functions = require('firebase-functions');

const onCreate = require('@background/onCreate');
const onDelete = require('@background/onDelete');
const onUpdate = require('@background/onUpdate');

const tours  = require('@http/tours');
const stripe = require('@http/stripe');

const docPath = 'tours/{tourId}';

module.exports.onCreate = functions.firestore.document(docPath).onCreate(onCreate);
module.exports.onDelete = functions.firestore.document(docPath).onDelete(onDelete);
module.exports.onUpdate = functions.firestore.document(docPath).onUpdate(onUpdate);

module.exports.tours  = functions.https.onRequest(tours);
module.exports.stripe = functions.https.onRequest(stripe);

CI / CD

每次将更改推送到回购时都进行持续集成和部署如何?你可以使用谷歌谷歌云构建。它是免费的,直到某个点:)检查这个链接。

。/ cloudbuild.yaml

steps:
  - name: "gcr.io/cloud-builders/npm"
    args: ["run", "install:functions"]
  - name: "gcr.io/cloud-builders/npm"
    args: ["test"]
  - name: "gcr.io/${PROJECT_ID}/firebase"
    args:
      [
        "deploy",
        "--only",
        "functions",
        "-P",
        "${PROJECT_ID}",
        "--token",
        "${_FIREBASE_TOKEN}"
      ]

substitutions:
    _FIREBASE_TOKEN: nothing

其他回答

为了保持简单(但能完成工作),我个人是这样构造我的代码的。

布局

├── /src/                      
│   ├── index.ts               
│   ├── foo.ts           
│   ├── bar.ts
|   ├── db.ts           
└── package.json  

foo.ts

import * as functions from 'firebase-functions';
export const fooFunction = functions.database()......... {
    //do your function.
}

export const someOtherFunction = functions.database().......... {
    // do the thing.
}

bar.ts

import * as functions from 'firebase-functions';
export const barFunction = functions.database()......... {
    //do your function.
}

export const anotherFunction = functions.database().......... {
    // do the thing.
}

db.ts

import * as admin from 'firebase-admin';
import * as functions from 'firebase-functions';

export const firestore = admin.firestore();
export const realtimeDb = admin.database();

index.ts

import * as admin from 'firebase-admin';
import * as functions from 'firebase-functions';

admin.initializeApp(functions.config().firebase);
// above codes only needed if you use firebase admin

export * from './foo';
export * from './bar';

适用于任何嵌套级别的目录。也只需遵循目录中的模式即可。

这要归功于@zaidfazil的答案

Node 8 LTS现在可以与Cloud/Firebase函数一起使用,您可以使用扩展操作符执行以下操作:

/ package.json

"engines": {
  "node": "8"
},

/ index.js

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

module.exports = {
  ...require("./lib/foo.js"),
  // ...require("./lib/bar.js") // add as many as you like
};

/lib/foo.js

const functions = require("firebase-functions");
const admin = require("firebase-admin");

exports.fooHandler = functions.database
  .ref("/food/{id}")
  .onCreate((snap, context) => {
    let id = context.params["id"];

    return admin
      .database()
      .ref(`/bar/${id}`)
      .set(true);
  });

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);
}

我使用一个普通的JS引导加载器来自动包含我想使用的所有函数。

├── /functions
│   ├── /test/
│   │   ├── testA.js
│   │   └── testB.js
│   ├── index.js
│   └── package.json

index.js(引导)

/**
 * The bootloader reads all directories (single level, NOT recursively)
 * to include all known functions.
 */
const functions = require('firebase-functions');
const fs = require('fs')
const path = require('path')

fs.readdirSync(process.cwd()).forEach(location => {
  if (!location.startsWith('.')) {
    location = path.resolve(location)

    if (fs.statSync(location).isDirectory() && path.dirname(location).toLowerCase() !== 'node_modules') {
      fs.readdirSync(location).forEach(filepath => {
        filepath = path.join(location, filepath)

        if (fs.statSync(filepath).isFile() && path.extname(filepath).toLowerCase() === '.js') {
          Object.assign(exports, require(filepath))
        }
      })
    }
  }
})

这个例子index.js文件只在根目录中自动包含目录。它可以扩展到walk目录,honor .gitignore等。不过这对我来说已经足够了。

有了索引文件,添加新函数就很简单了。

/测试/ testA.js

const functions = require('firebase-functions');

exports.helloWorld = functions.https.onRequest((request, response) => {
 response.send("Hello from Firebase!");
});

/测试/ testB.js

const functions = require('firebase-functions');

exports.helloWorld2 = functions.https.onRequest((request, response) => {
 response.send("Hello again, from Firebase!");
});

NPM运行服务产生:

λ ~/Workspace/Ventures/Author.io/Firebase/functions/ npm run serve

> functions@ serve /Users/cbutler/Workspace/Ventures/Author.io/Firebase/functions
> firebase serve --only functions


=== Serving from '/Users/cbutler/Workspace/Ventures/Author.io/Firebase'...

i  functions: Preparing to emulate functions.
Warning: You're using Node.js v9.3.0 but Google Cloud Functions only supports v6.11.5.
✔  functions: helloWorld: http://localhost:5000/authorio-ecorventures/us-central1/helloWorld
✔  functions: helloWorld2: http://localhost:5000/authorio-ecorventures/us-central1/helloWorld2

这个工作流程基本上就是“编写并运行”,而不必在每次添加/修改/删除新函数/文件时修改index.js文件。

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;