title: Facebook Messenger Chatbot messages tagging (of sorts)
body: <p>I have a Facebook Messenger bot (written in NodeJS) and a separate control panel where a user can manage the information that the bot is working with (like inventory stock, etc.). One of those things is a log of all conversations between the bot and a visitor. The control panel allows the admin users to send messages to visitors through the bot. There is an input box where they can type in a message and when they click 'Send', the message goes to the bot app, which then sends it back to the user through Send API.</p>
  
  <p>Messages are logged into a database; those going to the bot (from the visitor) are logged when they're received, and those the bot responds with are logged through the 'echo' callback.</p>
  
  <p>The problem with this is that the bot can reply to certain visitor commands (phrases) and tries to perform certain actions based on the input. I'm using Wit.ai for this, but due to the scope of the possible phrases and keywords, the default mode when someone sends a text message is to send it to Wit.ai for processing. However, if an admin user sends the visitor a message from the control panel, the visitor could want to respond to that message (instead of sending a bot command) and that response should not go to the Wit.ai for processing. And due to the sheer scope of possible variations of what can be said, coupled with the fact that they can actually use some of the keywords in the response as well, processing the intent with Wit.ai in that case is too uncertain.</p>
  
  <p>I was wondering if there's a way to somehow identify/mark the source of the messages that the bot sends to the visitor, so when an echo callback comes, I can know if it's, say, from a regular bot routine or from a user-entered reply. Like some additional meta tags that could be sent with the message that would also get returned with the echo, but that doesn't pollute the message itself. Is something like that possible? Or is there a different way I can achieve the same result.</p>