回复线程 google-api-ruby-client

Question

这就是我的代码（差不多）使用 google-api-ruby-客户端创建消息的样子：

  service ||= Google::Apis::GmailV1::GmailService.new

  message = RMail::Message.new
  message.header['To'] = params[:gmail][:to]
  message.header['From'] = current_user_google_user_id
  message.header['Subject'] = params[:gmail][:subject]
  message.header['Subject'] = params[:gmail][:subject]
  message.body = params[:gmail][:body]

  service.send_user_message(
    current_user_google_user_id,
    upload_source: StringIO.new(message.to_s),
    content_type: 'message/rfc822',
    thread_id: params[:gmail][:thread_id]
  )

它显然失败了，因为我有 thread_id。如果我删除该行，一切正常，但我无法将事情限制在线程范围内。我应该如何将线程 ID 传递给 GmailService？

Answer 1

查看 GitHub 上的源代码，send_user_message shows that it doesn't take a thread_id as a parameter. However the Message class 确实将其作为一个属性。

所以也许尝试这个应该有效：

  service ||= Google::Apis::GmailV1::GmailService.new

  message = RMail::Message.new
  message.header['To'] = params[:gmail][:to]
  message.header['From'] = current_user_google_user_id
  message.header['Subject'] = params[:gmail][:subject]
  message.header['Subject'] = params[:gmail][:subject]
  message.body = params[:gmail][:body]
  message.thread_id = params[:gmail][:thread_id]

  service.send_user_message(
    current_user_google_user_id,
    upload_source: StringIO.new(message.to_s),
    content_type: 'message/rfc822'
  )

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