Question

文档（https://core.telegram.org/bots/api#editmessagetext）说我需要指定要编辑的邮件ID，但我不知道如何获取该邮件ID。

我已尝试将ID设置为变量：

import telepot
from telepot.namedTuple import InlineKeyboardMarkup, InlineKeyboardButton

messageEditID = bot.sendMessage(<location ID>, "Test", reply_markup=InlineKeyboardMarkup(inline_keyboard=[[InlineKeyboardButton(text="Testing", callback_data="test"]]))['message_id']

因为此类消息的POST数据是

{"message_id":1140126,"chat":{"title":"<location name>","type":"supergroup","id":<location ID>},"date":1477960655,"from":{"username":"<bot username>","first_name":"<bot name>","id":<bot ID>},"text":"Test"}

然后根据内联回复数据回调它

if msg['data'] == 'test':
    bot.editMessage(messageEditID, "Did it work?")

但它会抛出一个Exception并显示以下消息：

Traceback (most recent call last):
  File "/usr/local/lib/python3.5/dist-packages/telepot/__init__.py", line 738, in collector
    callback(item)
  File "testingStuff.py", line 150, in handle
    bot.editMessageText(messageEditID, "Does it work?")
  File "/usr/local/lib/python3.5/dist-packages/telepot/__init__.py", line 592, in editMessageText
    return self._api_request('editMessageText', _rectify(p))
  File "/usr/local/lib/python3.5/dist-packages/telepot/__init__.py", line 398, in _api_request
    return api.request((self._token, method, params, files), **kwargs)
  File "/usr/local/lib/python3.5/dist-packages/telepot/api.py", line 131, in request
    return _parse(r)
  File "/usr/local/lib/python3.5/dist-packages/telepot/api.py", line 126, in _parse
    raise exception.TelegramError(description, error_code, data)
telepot.exception.TelegramError: ('Bad Request: Wrong inline message identifier specified', 400, {'description': 'Bad Request: Wrong inline message identifier specified', 'ok': False, 'error_code': 400})

当我尝试根据响应编辑邮件时，它引发了同样的错误，

if msg['data'] == 'test':
    msgID = msg['message']['message_id']
    bot.editMessageText(msgID, "OBOY")

因为传入内联回复的数据是：

{'chat_instance': '112564336693044113',
 'data': 'test',
 'from': {'first_name': 'B(', 'id': <user ID>, 'username': '<username>'},
 'id': '852908411206280027',
 'message': {'chat': {'id': <chat ID>,
                      'title': 'We da bes',
                      'type': 'supergroup',
                      'username': '<chat username>'},
             'date': 1477961765,
             'from': {'first_name': 'Testing For James',
                      'id': <bot ID>,
                      'username': '<bot username>'},
             'message_id': 63180,
             'text': 'Test'}}

任何人都可以帮我弄清楚我哪里出错了吗？

Answer 1

更新：Telepot有一个message_identifier方法，您可以通过将sendMessage方法设置为变量然后调用message_identifier消息来调用

sent = bot.sendMessage(9999999, "Testing")
edited = telepot.message_identifier(sent)
bot.editMessageText(edited, "Tested")

Answer 2

这是一个不使用 telepot

编辑消息的简单示例

演示：Before edit After edited

from telegram.ext import Updater, CommandHandler, MessageHandler, CallbackContext
from telegram import Update
import logging

# Enable logging
logging.basicConfig(format='%(asctime)s - %(name)s - %(levelname)s - %(message)s',
                    level=logging.INFO)

logger = logging.getLogger(__name__)


def start(update: Update, context: CallbackContext) -> None:
    context.bot.send_message(chat_id=update.message.chat_id, text="hi")


def edit(update: Update, context: CallbackContext) -> None:
    context.bot.editMessageText(chat_id=update.message.chat_id,
                                message_id=update.message.reply_to_message.message_id, 
                                text="edited")


def main() -> None:
    updater = Updater("TOKEN")
    updater.dispatcher.add_handler(CommandHandler('start', start))
    updater.dispatcher.add_handler(CommandHandler('edit', edit))
    updater.start_polling()
    updater.idle()


if __name__ == '__main__':
    main()

Answer 3

编辑：在我的回答中，我说的是python-telegram-bot，而不是关于telepot所以这不相关，抱歉。

如果您使用带按钮的消息（称为“InlineKeyboard”），则只能使用编辑功能。这是example

如何使用Telegram的python bot API编辑已发送的消息？

3 个答案: