【发布时间】:2021-05-02 07:42:56
【问题描述】:
我必须向多个用户发送多条消息。一些消息应该“等待”,直到用户单击该消息下的按钮。它应该看起来像这样(来自另一个更一般的question):
>>> bot: 'Hello'
>>> bot: 'My name is %bot name%'
>>> bot: 'How are you?'
<<< button: 'Fine' # <-- this is button. Any message should not to be sent before the user clicks `Fine` button
>>> bot: 'Can we continue?' # <-- this message shows only after user clicks `Fine` button
<<< button: 'Yes' # <-- this is button. Any message should not to be sent before the user clicks `Yes` button
>>> bot: 'That\'s great!' # <-- this message shows only after user clicks `Yes` button
<<< button: 'Of course' # <-- this is button. Any message should not to be sent before the user clicks `Of course` button
>>> bot: 'See you next time' # <-- this message shows only after user clicks `Yes` button
我尝试使用生成器实现该行为,但没有成功。
现在我已准备好对管道进行硬编码,但我仍然无法执行此操作。
下面的最小工作示例:
import time
from dataclasses import dataclass
from typing import Any, Dict, List
from config import Config
import telebot
bot = telebot.TeleBot(Config().get_str('test_bot_token'))
users: List[int] = []
@bot.message_handler(commands=['start'])
def _start(message: telebot.types.Message):
users.append(message.chat.id)
@bot.message_handler(func=lambda x: 'exec' in x.text)
def _send_message(message: telebot.types.Message):
for i, user in enumerate(users):
try:
_send_message_to_user(user)
except telebot.apihelper.ApiTelegramException as telegramEx:
print(f'chat_id:{user}\t{telegramEx}')
def _send_message_to_user(user_id: int):
@bot.callback_query_handler(func=lambda x: 'step_1' in x.data)
def exec_step_2(query: telebot.types.CallbackQuery):
bot.edit_message_reply_markup(chat_id=user_id,
message_id=query.message.id,
reply_markup=None)
step_kb = telebot.types.InlineKeyboardMarkup()
step_kb.add(telebot.types.InlineKeyboardButton('Yes', callback_data='step_2'))
bot.send_message(chat_id=user_id,
text='Can we continue?',
reply_markup=step_kb)
time.sleep(1.25)
@bot.callback_query_handler(func=lambda x: 'step_2' in x.data)
def exec_step_2(query: telebot.types.CallbackQuery):
bot.edit_message_reply_markup(chat_id=user_id,
message_id=query.message.id,
reply_markup=None)
step_kb = telebot.types.InlineKeyboardMarkup()
step_kb.add(telebot.types.InlineKeyboardButton('Of course', callback_data='step_3'))
bot.send_message(chat_id=user_id,
text='That\'s great!',
reply_markup=step_kb)
time.sleep(1.25)
@bot.callback_query_handler(func=lambda x: 'step_3' in x.data)
def exec_step_2(query: telebot.types.CallbackQuery):
bot.edit_message_reply_markup(chat_id=user_id,
message_id=query.message.id,
reply_markup=None)
bot.send_message(chat_id=user_id,
text='See you next time')
time.sleep(1.25)
bot.send_message(chat_id=user_id, text='Hello')
time.sleep(1.25)
bot.send_message(chat_id=user_id, text='My name is %bot name%')
time.sleep(1.25)
kb = telebot.types.InlineKeyboardMarkup()
kb.add(telebot.types.InlineKeyboardButton(text='Fine', callback_data='step_1'))
bot.send_message(chat_id=user_id, text='How are you?', reply_markup=kb)
bot.polling()
当多个用户附加到机器人时会出现问题。第一个用户获得完整的管道,但第二个用户 - 没有收到任何消息。
最简单的复制方法:
- 不要启动机器人
- 向机器人发送
/start命令 - 向机器人发送两条
exec消息 - 启动机器人
你会得到两个你好,我的名字是%bot name%,你好吗?消息。如果您单击任何按钮 Fine(第一个或第二个),您将在聊天顶部看到 Loading... 消息,并且不会得到任何响应。
但如果您发送/start 命令和单个exec 消息,一切都会正常工作。
【问题讨论】:
标签: python python-multithreading py-telegram-bot-api