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?
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
└── package.json
foo.ts
export const fooFunction = functions.database()......... {
//do your function.
}
export const someOtherFunction = functions.database().......... {
// do the thing.
}
bar.ts
export const barFunction = functions.database()......... {
//do your function.
}
export const anotherFunction = functions.database().......... {
// do the thing.
}
index.ts
import * as fooFunctions from './foo';
import * as barFunctions from './bar';
module.exports = {
...fooFunctions,
...barFunctions,
};
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。
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);
});
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
└── package.json
foo.ts
export const fooFunction = functions.database()......... {
//do your function.
}
export const someOtherFunction = functions.database().......... {
// do the thing.
}
bar.ts
export const barFunction = functions.database()......... {
//do your function.
}
export const anotherFunction = functions.database().......... {
// do the thing.
}
index.ts
import * as fooFunctions from './foo';
import * as barFunctions from './bar';
module.exports = {
...fooFunctions,
...barFunctions,
};
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。
