How to stop "listening" to bot.*_handler?

Question

PHP

Good day everyone!
I'll tell you in advance I rummaged through the whole Google and could not find the answer. I am
writing my first telegram bot using
pyTelegrambotAPI

@bot.message_handler(commands=['rename'])
def rename (massage):
    bot.send_message(massage.from_user.id, f'Привет , твое имя в телеграмм: {massage.from_user.first_name} \n'
                                           f'На какое хотите заменить?')
    bot.register_next_step_handler(massage,save_name)
def save_name (massage):
    global user_name
    user_name = massage.text
    quest = f'хорош, тепеь я буду называть Вас {user_name}, сохраняем?'
    keybord = keybord_yes_or_no()
    bot.send_message(massage.from_user.id, text= quest, reply_markup=keybord)
@bot.callback_query_handler(func=lambda call: True )
def rename_seve(call):
    if call.data == 'yes':
        global users
        users = ({
            'id': call.from_user.id,
            'name': user_name
        })
        bot.send_message(call.message.chat.id, f"Теперь для Вашего id {users['id']} будет сохранено имя {users['name']}")
    elif call.data == 'no':
        bot.send_message(call.from_user.id, 'Ок ничего не сохраняю')

and let's say I call the keyboard elsewhere:

@bot.message_handler(func=lambda message: sp.spelled(message.text.lower()) == "привет")
def hi (massage):
    bot.send_message(massage.chat.id, f"Привет ")
    keybord_1 = keybord_yes_or_no()
    q= '>>>>>>'
    bot.send_message(massage.chat.id,text=q, reply_markup=keybord_1)
@bot.callback_query_handler(func=lambda call: True)
def requset(call):
    if call.data == 'yes':
        bot.send_message(call.from_user.id, '>>>>????')
    elif call.data == 'no':
        bot.send_message(call.chat.id, '>>>>>!!!')

And of course it "listens" to the first decorator

. Question:
How to stop listening to the first decorator or the function itself so that it goes to the second one and starts listening to input on it?

Answer 1

Well, the error is obviously here: R::store($user);
and to help solve it, you should throw off the code of this function.
And also read php.net/manual/ru/book.filter.php , it will make life easier

Answer 2

In this version, I think instead of decorators, you should use the familiar bot.register_next_step_handler(massage,save_name) method . In the example from the github , you can see how the question is what gender? and two buttons m and w are added , and in the message where they are sent, they are deleted, and the response is processed as text.

def process_age_step(message):
    try:
        chat_id = message.chat.id
        age = message.text
        if not age.isdigit():
            msg = bot.reply_to(message, 'Age should be a number. How old are you?')
            bot.register_next_step_handler(msg, process_age_step)
            return
        user = user_dict[chat_id]
        user.age = age
        markup = types.ReplyKeyboardMarkup(one_time_keyboard=True)
        markup.add('Male', 'Female')
        msg = bot.reply_to(message, 'What is your gender', reply_markup=markup)
        bot.register_next_step_handler(msg, process_sex_step)
    except Exception as e:
        bot.reply_to(message, 'oooops')


def process_sex_step(message):
    try:
        chat_id = message.chat.id
        sex = message.text

I would also like to suggest making changes to the code, in such cases:

keybord = keybord_yes_or_no()
    bot.send_message(massage.from_user.id, text= quest, reply_markup=keybord)

If you generate buttons in this way, there will be one line less:

def day_btns():
  days = types.InlineKeyboardMarkup(row_width=7)
  days.add(*[types.InlineKeyboardButton(text='Day '+str(i),callback_data='call'+str(i)) for i in range(1,8)])
  return days

#Пример использования
bot.send_message(message.chat.id,'Days of week',reply_markup=day_btns())
#Хватает и одной строки

Also, if you are interested , here is a complete sample of my test bot