python - 如何使用python从网站中提取所有链接

标签 python web-scraping

<分区>

我写了一个脚本来从网站中提取链接,效果很好 这是源码


import requests
from bs4 import BeautifulSoup
Web=requests.get("https://www.google.com/")
soup=BeautifulSoup(Web.text,'lxml')
for link in soup.findAll('a'):
    print(link['href'])

##Out put
https://www.google.com.sa/imghp?hl=ar&tab=wi
https://maps.google.com.sa/maps?hl=ar&tab=wl
https://www.youtube.com/?gl=SA&tab=w1
https://news.google.com/?tab=wn
https://mail.google.com/mail/?tab=wm
https://drive.google.com/?tab=wo
https://calendar.google.com/calendar?tab=wc
https://www.google.com.sa/intl/ar/about/products?tab=wh
http://www.google.com.sa/history/optout?hl=ar
/preferences?hl=ar
https://accounts.google.com/ServiceLogin?hl=ar&passive=true&continue=https://www.google.com/&ec=GAZAAQ
/search?safe=strict&ie=UTF-8&q=%D9%86%D9%88%D8%B1+%D8%A7%D9%84%D8%B4%D8%B1%D9%8A%D9%81&oi=ddle&ct=174786979&hl=ar&kgmid=/m/0562zv&sa=X&ved=0ahUKEwiq8feoiqDwAhUK8BQKHc7UD7oQPQgD
/advanced_search?hl=ar-SA&authuser=0
https://www.google.com/setprefs?sig=0_mwAqJUgnrqSouOmGk0UvVz7GgkY%3D&hl=en&source=homepage&sa=X&ved=0ahUKEwiq8feoiqDwAhUK8BQKHc7UD7oQ2ZgBCAU
/intl/ar/ads/
http://www.google.com/intl/ar/services/
/intl/ar/about.html
https://www.google.com/setprefdomain?prefdom=SA&prev=https://www.google.com.sa/&sig=K_e_0jdE_IjI-G5o1qMYziPpQwHgs%3D
/intl/ar/policies/privacy/
/intl/ar/policies/terms/

但问题是,当我将网站更改为 https://www.jarir.com/ 时, 它不会起作用

import requests
from bs4 import BeautifulSoup
Web=requests.get("https://www.jarir.com/")
soup=BeautifulSoup(Web.text,'lxml')
for link in soup.findAll('a'):
    print(link['href'])

#out put
#

输出将是#

最佳答案

我建议添加一个随机 header 函数以避免网站将 python-requests 检测为浏览器/代理。下面的代码按要求返回所有链接。

请注意 header 的随机化以及此代码如何在 requests.get 方法中使用 header 参数。

import requests
from bs4 import BeautifulSoup
from random import choice

desktop_agents = [
    'Mozilla/5.0 (Windows NT 6.1; WOW64) AppleWebKit/537.36 (KHTML, like Gecko) Chrome/54.0.2840.99 Safari/537.36',
    'Mozilla/5.0 (Windows NT 10.0; WOW64) AppleWebKit/537.36 (KHTML, like Gecko) Chrome/54.0.2840.99 Safari/537.36',
    'Mozilla/5.0 (Windows NT 10.0; Win64; x64) AppleWebKit/537.36 (KHTML, like Gecko) Chrome/54.0.2840.99 Safari/537.36',
    'Mozilla/5.0 (Macintosh; Intel Mac OS X 10_12_1) AppleWebKit/602.2.14 (KHTML, like Gecko) Version/10.0.1 Safari/602.2.14',
    'Mozilla/5.0 (Windows NT 10.0; WOW64) AppleWebKit/537.36 (KHTML, like Gecko) Chrome/54.0.2840.71 Safari/537.36',
    'Mozilla/5.0 (Macintosh; Intel Mac OS X 10_12_1) AppleWebKit/537.36 (KHTML, like Gecko) Chrome/54.0.2840.98 Safari/537.36',
    'Mozilla/5.0 (Macintosh; Intel Mac OS X 10_11_6) AppleWebKit/537.36 (KHTML, like Gecko) Chrome/54.0.2840.98 Safari/537.36',
    'Mozilla/5.0 (Windows NT 6.1; WOW64) AppleWebKit/537.36 (KHTML, like Gecko) Chrome/54.0.2840.71 Safari/537.36',
    'Mozilla/5.0 (Windows NT 6.1; Win64; x64) AppleWebKit/537.36 (KHTML, like Gecko) Chrome/54.0.2840.99 Safari/537.36',
    'Mozilla/5.0 (Windows NT 10.0; WOW64; rv:50.0) Gecko/20100101 Firefox/50.0']


def random_headers():
    return {'User-Agent': choice(desktop_agents),
            'Accept': 'text/html,application/xhtml+xml,application/xml;q=0.9,image/webp,*/*;q=0.8'}

Web=requests.get("https://www.jarir.com", headers=random_headers())
soup=BeautifulSoup(Web.text,'lxml')
for link in soup.findAll('a'):
    print(link['href'])

关于python - 如何使用python从网站中提取所有链接,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/67293733/

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