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?
org大纲是一个更简单的体系结构模式,可以将方法分离到不同的文件中,并在index.js文件中的一行中导出。
本示例中项目的架构如下:
projectDirectory
index.js
podcast.js
profile.js
index.js
const admin = require('firebase-admin');
const podcast = require('./podcast');
const profile = require('./profile');
admin.initializeApp();
exports.getPodcast = podcast.getPodcast();
exports.removeProfile = profile.removeProfile();
podcast.js
const functions = require('firebase-functions');
exports.getPodcast = () => functions.https.onCall(async (data, context) => {
...
return { ... }
});
同样的模式将用于概要文件中的removeProfile方法。
@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');
}
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);
});
有一种很好的方法可以长期组织所有的云功能。我最近就这么做了,效果完美无缺。
我所做的是根据每个云函数的触发端点将它们组织在单独的文件夹中。每个云函数的文件名都以*.f.js结尾。例如,如果你在user/{userId}/document/{documententid}上有onCreate和onUpdate触发器,那么在functions/user/document/目录下创建两个文件onCreate.f.js和onUpdate.f.js,你的函数将分别命名为userDocumentOnCreate和userDocumentOnUpdate。(1)
下面是一个目录结构示例:
functions/
|----package.json
|----index.js
/----user/
|-------onCreate.f.js
|-------onWrite.f.js
/-------document/
|------------onCreate.f.js
|------------onUpdate.f.js
/----books/
|-------onCreate.f.js
|-------onUpdate.f.js
|-------onDelete.f.js
样本函数
const functions = require('firebase-functions');
const admin = require('firebase-admin');
const db = admin.database();
const documentsOnCreate = functions.database
.ref('user/{userId}/document/{documentId}')
.onCreate((snap, context) => {
// your code goes here
});
exports = module.exports = documentsOnCreate;
Index.js
const glob = require("glob");
const camelCase = require('camelcase');
const admin = require('firebase-admin');
const serviceAccount = require('./path/to/ServiceAccountKey.json');
try {
admin.initializeApp({ credential: admin.credential.cert(serviceAccount),
databaseURL: "Your database URL" });
} catch (e) {
console.log(e);
}
const files = glob.sync('./**/*.f.js', { cwd: __dirname });
for (let f = 0, fl = files.length; f < fl; f++) {
const file = files[f];
const functionName = camelCase(file.slice(0, -5).split('/'));
if (!process.env.FUNCTION_NAME || process.env.FUNCTION_NAME === functionName) {
exports[functionName] = require(file);
}
}
你可以使用任何你想要的名字。对我来说,onCreate.f.js, onUpdate.f.js等似乎更相关的类型的触发器。
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
在我努力实现@zaidfazil的解决方案时,我想出了以下方法(使用JavaScript,而不是TypeScript)。
multi.js
exports.onQuestionMultiCreate = functions.database
.ref("/questions-multi/{questionId}")
.onCreate(async (snapshot, context) => {
...
}
});
trueFalse.js
exports.onQuestionTrueFalseCreate = functions.database
.ref("/questions-truefalse/{questionId}")
.onCreate(async (snapshot, context) => {
...
}
});
index.js
const multi = require("./multi");
const trueFalse = require("./trueFalse");
module.exports = {
...multi,
...trueFalse