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 云功能对存储桶中的视频进行转码，并使用在线发布的代码输出到另一个桶中，但进行了一些调整。

但我收到以下错误：

TypeError：无法在 transcodeVideo 读取未定义的属性“名称” (/srv/index.js:17:56) 在 /worker/worker.js:825:24 在 process._tickDomainCallback (internal/process/next_tick.js:229:7)

Index.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();

  // Transcode
  ffmpeg()
    .input(remoteReadStream)
    .outputOptions('-c:v libx264') // Change these options to whatever suits your needs
    .outputOptions('-c:a copy')
    .outputOptions('-vf "scale=4800:2400"')
    .outputOptions('-b:a 160k')
    .outputOptions('-b:a 160k')
    .outputOptions('-x264-params mvrange=511')
    .outputOptions('-f mp4')
    .outputOptions('-preset slow')
    .outputOptions('-movflags frag_keyframe+empty_moov')
    .outputOptions('-crf 18')
    // https://github.com/fluent-ffmpeg/node-fluent-ffmpeg/issues/346#issuecomment-67299526 
    .on('start', (cmdLine) => {
      console.log('Started ffmpeg with command:', cmdLine);
    })
    .on('end', () => {
      console.log('Successfully re-encoded video.');
      callback();
    })
    .on('error', (err, stdout, stderr) => {
      console.error('An error occured during encoding', err.message);
      console.error('stdout:', stdout);
      console.error('stderr:', stderr);
      callback(err);
    })
    .pipe(remoteWriteStream, { end: true }); // end: true, emit end event when readable stream ends
};

Answer 1

函数transcodeVideo使用参数（事件、回调）获取文件数据的问题。根据document，这些参数（事件和回调）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node -v 检查 Node.js 版本，或者在控制台的 Cloud Function 部分中检查它。 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。