云功能:如何上传其他文件以在代码中使用? [英] Cloud Functions: how to upload additional file for use in code?

问题描述

我需要访问我的代码中的protoc文件.在本地，我只是将其放在文件夹中，但是如何从已部署的Firebase函数中获取此文件?

I need to get access to protoc file in my code. Locally I just put it in the folder but how to get this file from deployed Firebase functions?

const grpc = require('grpc');
const PROTO_PATH = __dirname + '\\protos\\prediction_service.proto';

exports.helloWorld = functions.https.onRequest((request, response){
    var tensorflow_serving = grpc.load(PROTO_PATH).tensorflow.serving;
...
}

推荐答案

您要上传3个文件来部署您的Cloud Function:

You'd like to upload 3 files to deploy your Cloud Function:

• index.js
• package.json
• prediction_service.proto

要通过开发者控制台执行此操作，您需要:

In order to do so via the Developer Console, you'll need to:

1. 转到Google Cloud开发者控制台> Cloud Functions>创建函数
2. 在源代码"字段中，选择以下任一项:
   • "ZIP上传"，然后选择一个包含3个文件的zip文件，
   • 来自云存储的ZIP"并选择GCS上的文件位置，
   • "Cloud Source存储库"并提供您的回购详细信息

1. Go to the Google Cloud Developer Console > Cloud Functions > Create Function
2. In the "Source Code" field, choose either:
   • "ZIP upload" and select a zip file including your 3 files,
   • "ZIP from Cloud Storage" and select file location on GCS,
   • "Cloud Source repository" and provide your repo details

部署后，在功能的源"选项卡中，您将看到显示的三个文件.

Once deployed, in the source tab for your function you'll see the three files displayed.

或者，您可以使用gcloud通过以下命令来部署文件:

Alternatively, you can use gcloud to deploy your files via the following command:

gcloud beta functions deploy <functionName> --source=SOURCE

其中源可以是Google Cloud Storage上的ZIP文件，对源存储库的引用或本地文件系统路径.我建议您查看 doc 此命令以获取全部详细信息.

where source can be a ZIP file on Google Cloud Storage, a reference to source repository or a local filesystem path. I'd recommend to have a look at the doc for this command for full details.

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