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

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) => {
    ...
};

其他回答

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) => {
    ...
};

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;

@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'); 
}

我花了很多时间寻找同样的东西,我认为这是实现它的最佳方法(我使用firebase@7.3.0):

https://codeburst.io/organizing-your-firebase-cloud-functions-67dc17b3b0da

毫不费力;)

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);
  });