小编Joe*_*Joe的帖子

无需用户输入即可向电报组发送消息

我正在尝试构建一个机器人,它会在使用 python 的最新新闻中有更新时自动发送消息。以下是我所做的。

companies = {
    "name_1": {
        "rss": "name_1 rss link",
        "link": "name_1 link"
    }
}

import feedparser as fp
import time, telebot
token = <TOKEN>
bot = telebot.TeleBot(token)
LIMIT = 1
while True:
    def get_news():
        count = 1
        news = []
        for company, value in companies.items():
            count = 1
            if 'rss' in value:
                d = fp.parse(value['rss'])
                for entry in d.entries:
                    if hasattr(entry, 'published'):
                        if count > LIMIT:
                            break
                        news.append(entry.link)
                        count = count + 1

        return (news)
    val = …
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python-3.x telegram python-telegram-bot telegram-bot

6
推荐指数
2
解决办法
2万
查看次数

Telegram bot:如何在按钮点击时要求用户输入

我一直在尝试创建一个电报机器人。我用 Python 编码,以下是我到目前为止所做的。

import logging
from telegram import InlineKeyboardButton, InlineKeyboardMarkup
from telegram.ext import Updater, CommandHandler, CallbackQueryHandler

logging.basicConfig(format='%(asctime)s - %(name)s - %(levelname)s - %(message)s',
                    level=logging.INFO)
logger = logging.getLogger(__name__)

@app.route("/")
def start(bot, update):
    help_text = 'text'
    keyboard = [[InlineKeyboardButton("Button_1", url=('url'))]]
    reply_markup = InlineKeyboardMarkup(keyboard)
    update.message.reply_text(help_text, reply_markup=reply_markup)


def button(bot, update):
    query = update.callback_query
    bot.edit_message_text(text="Selected option: {}".format(query.data),
                          chat_id=query.message.chat_id,
                          message_id=query.message.message_id)

def error(bot, update, error):
    logger.warning('Update "%s" caused error "%s"', update, error)


def main():
    updater = Updater(<TOKEN>)
    updater.dispatcher.add_handler(CommandHandler('start', start))
    updater.dispatcher.add_handler(CallbackQueryHandler(button))
    updater.dispatcher.add_error_handler(error)
    updater.start_polling()
    updater.idle()


if __name__ == "__main__": …
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python-3.x python-telegram-bot telegram-bot

5
推荐指数
1
解决办法
435
查看次数

在python中以一定间隔从右边拆分一个字符串

我正试图从右边分割一个字符串.以下是代码.

string = "abcde" 
n = len(string)
slices = [string[i-3:i] for i in range(n,0,-3)]
print (slices)
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我把输出作为['cde', ''].我想要的['cde', 'ab']

但是当我从左边分开它时,它给出了正确的输出,即,

string = "abcde" 
slices = [string[i:i+3] for i in range(0,n,3)]
print (slices)

output: ['abc', 'de']
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任何人都可以指出我哪里出错了?

python string split list

4
推荐指数
1
解决办法
109
查看次数