Is there a way to send a message (call a function), without a command from the bot's user? For example when a certain condition is met?
Currently I have only been able to call a function with a command.
I have been trying the following:
from telegram.ext import Updater, CommandHandler, CallbackContext
from telegram import Update
from datetime import datetime
import telebot as tb
import logging
bot = tb.TeleBot(API_KEY)
logging.basicConfig(
format='%(asctime)s - %(name)s - %(levelname)s - %(message)s', level=logging.INFO
)
logger = logging.getLogger(__name__)
def messager(update: Update, context: CallbackContext) -> None:
"""this function messages at a specific time"""
update.message.reply_text("message back")
if __name__ == "__main__":
updater = Updater(API_KEY)
dispatcher = updater.dispatcher
dispatcher.add_handler(CommandHandler("message", messager))
condition = False #I have been changing this when testing
if (condition == True):
messager() # how to call messager here? how to pass update and context arguments?
updater.start_polling()
updater.idle()
Any possible way to achieve this would be great.
You cannot call a function using update
and context
since there is no update from the telegram servers to act on. In other words there is no event to perform a callback on.
Depending on what you want to do you could use a MessageHandler
instead of a CommandHandler
to reply to any message with Filters
to filter out which ones to reply to
or if you want to send a message every time you start your bot and you know who to send a message to (ie you have their chat_id
) you can do
if condition:
updater.bot.send_message(chat_id=CHAT_ID, text="message back")
or if you just want to send a message every time some user starts a chat with your bot you could use the start
command which is sent by default to start a chat with a bot
The technical post webpages of this site follow the CC BY-SA 4.0 protocol. If you need to reprint, please indicate the site URL or the original address.Any question please contact:yoyou2525@163.com.