[XmlHttpRequest] xhr.upload.onprogress not working with xhr.send({uri: 'file:///...'}) #15724

var xhr = new XMLHttpRequest();
xhr.open('PUT', this.url, true);
xhr.setRequestHeader('Content-Type', this.contentType); <=== video/mp4
xhr.setRequestHeader('Content-Range', 'bytes ' + this.offset + '-' + (end - 1) + '/' + this.filesize);
xhr.setRequestHeader('X-Upload-Content-Type', this.filetype);  <=== video/mp4

if (xhr.upload) {
  xhr.upload.addEventListener('progress', this.onProgress);
}
xhr.onload = this.onContentUploadSuccess_.bind(this);
xhr.onerror = this.onContentUploadError_.bind(this);
xhr.send({uri: this.file.path});  <=== file:///.....

let data = new FormData();
data.append('File', {
      uri:  source.path,
      type: source.type,
      name: source.name
});
var xhr = new XMLHttpRequest();
  xhr.open('PUT', this.url, true);
  //xhr.setRequestHeader('Content-Type', this.contentType);  
  //=== comment this, because it throw error 'multipart != video/mp4'
  //xhr.setRequestHeader('Content-Range', 'bytes ' + this.offset + '-' + (end - 1) + '/' + this.filesize);
  //=== comment this, because sending data size != filesize
  xhr.setRequestHeader('X-Upload-Content-Type', this.filetype);

  if (xhr.upload) {
    xhr.upload.addEventListener('progress', this.onProgress);
  }
  xhr.onload = this.onContentUploadSuccess_.bind(this);
  xhr.onerror = this.onContentUploadError_.bind(this);
  xhr.send(data); <==== FormData