获取Instagram个人资料的名称和与Python的发布日期



我正在学习python3的过程中,我尝试解决一个简单的任务。我想从Instagram链接中获取帐户的名称和post 的

import requests
from bs4 import BeautifulSoup
html = requests.get('https://www.instagram.com/p/BuPSnoTlvTR')
soup = BeautifulSoup(html.text, 'lxml')
item = soup.select_one("meta[property='og:description']")
name = item.find_previous_sibling().get("content").split("•")[0]
print(name)

此代码有时与此类链接一起使用https://www.instagram.com/ yem> kingtop 但是我还需要它也可以与https://www.instagram.com/p/buxb00kfi-x/

这样的图像帖子使用

这就是我所能做的,但这是不起作用的。而且我也无法约会。你有什么想法?感谢任何帮助。

我找到了获取帐户名称的方法。现在,我正在尝试找到一种获得上传日期的方法

import requests
from bs4 import BeautifulSoup
import urllib.request
import urllib.error
import time
from multiprocessing import Pool
from requests.exceptions import HTTPError
start = time.time()
file = open('users.txt', 'r', encoding="ISO-8859-1")
urls = file.readlines()
for url in urls:
url = url.strip ('n')
try:
    req = requests.get(url)
    req.raise_for_status()
except HTTPError as http_err:
    output = open('output2.txt', 'a')
    output.write(f'не найденаn')  
except Exception as err:
    output = open('output2.txt', 'a')
    output.write(f'не найденыn')  
else:
    output = open('output2.txt', 'a')
    soup = BeautifulSoup(req.text, "lxml")
    the_url = soup.select("[rel='canonical']")[0]['href']
    the_url2=the_url.replace('https://www.instagram.com/','')
    head, sep, tail = the_url2.partition('/')
    output.write (head+'n')

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