Pytelegrambotapi直列Google搜索引擎
@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)
当我编写@bot Google请求时
机器人将其视为G GO GOOGEE
是什么导致错误“对电报API的请求不成功。错误代码：400。描述：不良请求：查询太旧了，响应超时过期或查询ID无效”
如何制作文本输入超时，以免对每个字母响应？
@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?
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评论（1）
我认为，错误在于您解析数据的方式。为了获得答案方法，至少需要8秒钟（基于睡眠）。电报内联查询只有几秒钟的时间，直到被认为是旧的，因此，在调用
bot.answer_inline_query（）之后，最好处理数据，然后使用bot.send_message（ ）我不确定它如何与异步代码一起使用。
如果您找到了另一个解决方案，请告诉我:)
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 usingbot.send_message()I am not certain how it works with async code though.
If you find another solution, please let me know :)