谷歌函数：TypeError：无法读取未定义的属性“名称”

Question

我正在尝试使用 Google 云功能将视频从存储桶转码，并使用在线发布的代码输出到另一个存储桶，但进行了一些调整。

但我收到以下错误：

类型错误：无法读取在 transcodeVideo (/srv/index.js:17:56) at /worker/worker.js:825:24 at at process._tickDomainCallback (internal/process/next_tick.js:229) 处未定义的属性“名称” :7)

索引.js

const {Storage} = require('@google-cloud/storage');

const projectId = 'cc18-223318';

const storage = new Storage({

    projectId: projectId,

});

const ffmpegPath = require('@ffmpeg-installer/ffmpeg').path;

const ffmpeg = require('fluent-ffmpeg');

const transcodedBucket = storage.bucket('2400p');

const uploadBucket = storage.bucket('inpuut');

ffmpeg.setFfmpegPath(ffmpegPath);

exports.transcodeVideo = function transcodeVideo(event, callback) {

  const file = event.data;

  // Ensure that you only proceed if the file is newly created, and exists.

  if (file.metageneration !== '1' || file.resourceState !== 'exists') {

    callback();

    return;

  }

  // Open write stream to new bucket, modify the filename as needed.

  const remoteWriteStream = transcodedBucket.file(file.name.replace('.webm', '.mp4'))

    .createWriteStream({

      metadata: {

        metadata: file.metadata, // You may not need this, my uploads have associated metadata

        contentType: 'video/mp4', // This could be whatever else you are transcoding to

      },

    });

  // Open read stream to our uploaded file

  const remoteReadStream = uploadBucket.file(file.name).createReadStream();

Answer 1

函数 transcodeVideo 使用参数（事件、回调）获取文件数据的问题。这些参数（事件和回调）根据本文档用于 Node.js 6 ，对于Node.js 8/10，您应该使用（数据和上下文）。请使用命令检查 Node.js 版本node -v，或者只是在控制台的 Cloud Function 部分检查它。解决方案是安装旧的 Node.js 6 版本或编辑与 Node.js 8/10 兼容的代码。