@bot.inline_handler(func=lambda query: len(query.query) > 0)
def query_text(query):
sleep(6)
text=query.query
html=requests.get(f'https://google.com/search?q={text}')
# print(html.status_code)
open('index.html','w', encoding='utf-8').write(html.text)
soup=BeautifulSoup(html.text, 'html.parser').find_all('div',{"class":"***********"})
for i in soup:
fk.append(types.InlineQueryResultArticle(id=str(len(fk)), title=f"{i.find('h3').get_text()}",description=f"{i.find('div',{'class':'**********'}).get_text()}",input_message_content=types.InputTextMessageContent(message_text=i.find('a').get('href').replace('/url?q=','https://google.com/url?q=')),hide_url=True,url=i.find('a').get('href').replace('/url?q=','https://google.com/url?q='),thumb_url='https://w7.pngwing.com/pngs/338/520/png-transparent-g-suite-google-play-google-logo-google-text-logo-cloud-computing.png', thumb_width=30, thumb_height=30))
print(i.find('a').get('href').replace('/url?q=','')+'\n')
sleep(2)
bot.answer_inline_query(query.id, fk)
When I write @bot google request
Bot takes it as g go goo google
What is causing the error
"A request to the Telegram API was unsuccessful. Error code: 400. Description: Bad Request: query is too old and response timeout expired or query ID is invalid"
How to make text input timeout so that it doesn't respond to every letter?
I think, the error resides in your way of parsing data. It takes at least 8 seconds (based on sleeps) just to get to the answer method. Telegram inline queries have very few seconds until they are considered old, so, it is better to process data after you call bot.answer_inline_query() and then send it to user using bot.send_message()
I am not certain how it works with async code though. If you find another solution, please let me know :)