我正试图每x秒发送一个InlineKeyboardHandler
。为此,我使用了updater.job_queue.run_repeating
,但它的行为很奇怪
除非我先与机器人进行另一次交互,否则键盘无法工作。我已经编写了一段简单的代码,您可以进行测试
from telegram import Update, InlineKeyboardButton, InlineKeyboardMarkup
from telegram.ext import Updater, CommandHandler, ConversationHandler, CallbackContext, CallbackQueryHandler
user_id = '*********'
tlg_token = '******************************'
SELECTING_COMMAND=1
keyboard = [[InlineKeyboardButton('Button: Print Clicked', callback_data=1)],]
reply_markup = InlineKeyboardMarkup(keyboard)
def menu(update: Update, context: CallbackContext) -> int:
update.message.reply_text('sent by command button:', reply_markup=reply_markup)
return SELECTING_COMMAND
def InlineKeyboardHandler(update: Update, _: CallbackContext) -> None:
print('clicked')
return 1
def cancel(update: Update, context: CallbackContext) -> int:
return ConversationHandler.END
updater = Updater(tlg_token, use_context=True)
dispatcher = updater.dispatcher
conv_handler = ConversationHandler(
entry_points=[CommandHandler('request_button', menu)],
states={
SELECTING_COMMAND: [CallbackQueryHandler(InlineKeyboardHandler)],
},
fallbacks=[CommandHandler('cancel', cancel)],
)
dispatcher.add_handler(conv_handler)
j = updater.job_queue
def talker(update):
update.bot.sendMessage(chat_id=user_id, text='sent by talker:', reply_markup=reply_markup)
j.run_repeating(talker, interval=10, first=0)
updater.start_polling()
updater.bot.sendMessage(chat_id=user_id, text='/request_button')
updater.idle()
我希望在单击按钮后可以看到“clicked”(单击)打印,但除非您先单击/request_按钮,否则它将无法工作。为什么?我怎样才能修好它
注释中提到的代码a_guest的问题是
InlineKeyboardHandler
只有在调用request_button
命令后才能开始工作下面是一个独立注册
InlineKeyboardHandler
的工作版本:该问题的另一个解决方案是OP在评论中提到的,您在评论中添加了CallbackQueryHandler作为入口点:
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