This code sends a message to the Telegram Supergroup if a new member has joined. When an error occurs when sending a message, I want to change my account to continue. It is possible to go to the next "item". How do I go to the next account in a loop when I receive an error?
from pyrogram import Client, Filters list_account = ['001', '002'] for item in list_account: app = Client(item) @app.on_message(Filters.chat("public_link_chat") & Filters.new_chat_members) def welcome(client, message): try: client.send_message( message.chat.id, 'Test', reply_to_message_id=message.message_id, disable_web_page_preview=True ) except Exception as e: print(e) # How do I go to the next account in a loop when I receive an error? app.start() app.join_chat("public_link_chat") app.idle() Function "continue" does not work in this case.
Description of function here: https://docs.pyrogram.ml/resources/UpdateHandling
1 Answers
Answers 1
Just add app.is_idle = False:
from pyrogram import Client, Filters list_account = ['001', '002'] for item in list_account: app = Client(item) @app.on_message(Filters.chat("public_link_chat") & Filters.new_chat_members) def welcome(client, message): try: client.send_message( message.chat.id, 'Test', reply_to_message_id=message.message_id, disable_web_page_preview=True ) except Exception as e: print(e) # How do I go to the next account in a loop when I receive an error? app.is_idle = False app.start() app.join_chat("public_link_chat") app.idle() You should definitely check out these lines of the idle logic at the pyrogram source code:
while self.is_idle: time.sleep(1) If you want an infinite loop, check out the itertools.cycle, it may be used like:
for item in itertools.cycle(list_account): do_something() 
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