python获取百度热榜链接的实例方法

https://www.baidu.com/

import requests
import pprint
import re
import urllib.parse

url = 'https://www.baidu.com/'

headers = {
    'Host': 'www.baidu.com',
    'Referer': 'https://www.baidu.com/',
    'User-Agent': 你的User-Agent,
    'Cookie': 你的Cookie
}

response = requests.get(url, headers=headers).content.decode('utf-8')
# 获取关键字
pat = '"pure_title": "(.*?)"'
keyword = re.findall(pat, response, re.S)
print(len(keyword))

for hot_word in keyword:
    # 汉字不符合url标准，所以这里需要进行url编码
    i = urllib.parse.quote(hot_word, encoding='utf-8', errors='replace')
    # url构建
    link = f'https://www.baidu.com/s?cl=3&tn=baidutop10&fr=top1000&wd={i}&rsv_idx=2&rsv_dl=fyb_n_homepage&hisfilter=1'
    print(link)

from lxml import etree
import re
import requests
import string
import json
headers = {
  "User-Agent":"Mozilla/5.0 (Windows NT 10.0; Win64; x64) AppleWebKit/537.36 (KHTML, like Gecko) Chrome/68.0.3440.106 Safari/537.36"
}
response = requests.get('https://www.baidu.com/s?wd=腾讯视频优惠&lm=1',headers=headers)
r = response.text
html = etree.HTML(r,etree.HTMLParser())
r1 = html.xpath('//h3')
r2 = html.xpath('//*[@class="c-abstract"]')
r3 = html.xpath('//a[@class="c-showurl"]')
for i in range(10) :
  r11 = r1[i].xpath('string(.)')
  r22 = r2[i].xpath('string(.)')
  r33 = r3[i].xpath('string(.)')
  # with open('test.txt', 'a', encoding='utf-8') as f:
  #   f.write(json.dumps(r11,ensure_ascii=False) + '\n')
  #   f.write(json.dumps(r22, ensure_ascii=False) + '\n')
  #   f.write(json.dumps(r33, ensure_ascii=False) + '\n')
  print(r11,end='\n')
  print(r22,end='\n')
  print(r33)
  print()