数据库触发firebase函数从URL下载图像并保存到存储区

Question

当我的数据库用“photo_url”字段更新时，我想下载图像并将其保存到存储区

exports.saveToStorage = functions.database.ref(`/images/${itemImageRef}`)
.onWrite(event => {
  const filePath = event.data.val();
  const filename = filePath.split('/').pop();

  var download = request.get(filePath).on('error', (err) => {
    console.log(err)
  })
  .pipe(fs.createWriteStream(filename));

  download.on('finish', () => {
    const bucket = gcs.bucket('id.appspot.com');
    const storagePath = `images/${filename}`;

    return bucket.upload(download, { destination: storagePath })
    .then(() => {
      console.log('success upload');
    });
  });
});

Answer 1

有了@jobsamuel建议的帖子，代码现在可以工作了：

exports.saveToStorage = functions.database.ref(`/images/${itemImageRef}`)
.onWrite(event => {

  const filePath = event.data.val();
  const filename = filePath.split('/').pop();

  const bucket = gcs.bucket('id.appspot.com');
  const remoteWriteStream = bucket.file(filename).createWriteStream({
    metadata: { contentType: 'image/jpeg' }
  });

  request(filePath).pipe(remoteWriteStream)
  .on('error', (err) => console.log(err))
  .on('finish', () => console.log('success save image'));

});

通过将请求结果直接管道到bucket，它通过跳过写入本地文件的步骤来解决问题，我怀疑这是我的原始代码失败的原因。另外，不要忘记为图像设置contentType。