Question

我有一个代码可以在instagram链接的URL中搜索，即使我可以在URL的源代码中手动找到，也不会返回一些URL。该URL没有被阻止，因为我可以在相同的URL中找到Facebook链接，但在instagram的URL中找不到。

import re
import requests

try:
    url = "https://www.walmart.com.br"
    headers = {'User-Agent': 'Mozilla/5.0 (Macintosh; Intel Mac OS X 10_10_1) AppleWebKit/537.36 (KHTML, like Gecko) Chrome/39.0.2171.95 Safari/537.36'}
    page = requests.get(url, headers = headers)
    page = page.text
    instagram_link = re.findall('https?://www.instagram.com/(?:[-\w.]|(?:%[\da-fA-F]{2}))+', page)[0]
    print("Instagram:", instagram_link)

except:
    pass
    instagram_link = "Not found"
    print("Instagram:", instagram_link)


try:
    facebook_link = re.findall('https?://www.facebook.com/(?:[-\w.]|(?:%[\da-fA-F]{2}))+', page)[0]
    print("Facebook:", facebook_link)

except:
    pass
    facebook_link = "Not found"      
    print("Facebook:", facebook_link)

>>> Instagram: Not found
>>> Facebook: https://www.facebook.com/walmart.com

Answer 1

instagram_link = re.findall('https?://www.instagram.com/(?:[-\w.]|(?:%[\da-fA-F]{2}))+', page)[0]

您在Instagram网址中忘记了www

我需要找出为什么有时不重新查找工作？

1 个答案: