Skype4Py: getting the handle of the person you're sending a message to

Question

I'm making a little bot for skype using skype4py. The bot is attached to MY skype client, and what I want is for the bot to accept input from either myself, or whoever i'm chatting with and to output to the chat window from where the input came.

To do this I use sendMessage(msgtext, handle_of_person_im_chatting_with).

It's pretty easy to accomplish when I receive a command from the person i'm chatting with because the RECEIVED event comes with a 'fromHandle' property containing their username, so I can direct sendMessage to that handle. But I can't figure out for the life of me how I could get that userhandle from the SENT event.

Answer 1

put this at the top after the imports

def commands(Message, Status):
    if Status == 'RECEIVED' or (Status == 'SENT'):
        if Message.Body.startswith(".help"):
            cmd_help(Message)

then for it to have an output for when .help is used put this

def cmd_commandhere(Message):
    Message.Chat.SendMessage("Hello, " + Message.Sender.Handle)
    print "commandhere command received from: " + Message.Sender.Handle

then at the bottom put this

skype = Skype4Py.Skype()
client = Skype4Py.client.Client(skype)
skype.OnMessageStatus = commands

Skype.Attach()

while True:
    raw_input('')