pyhton爬虫（简单过字体加密）

1. 找到反爬内容确定字体存放位置

定位到正则匹配

import re
from fontTools.ttLib import TTFont
import requests
headers = {
          
   
       "User-Agent": "Mozilla/5.0 (Windows NT 10.0; Win64; x64) AppleWebKit/537.36 (KHTML, like Gecko) Chrome/101.0.4951.54 Safari/537.36 Edg/101.0.1210.39"
}
url= https://www.maoyan.com/board/1?timeStamp=1652605412237&channelId=40011&index=8&signKey=bace7e65a24e717ead8cf5d252c0a1ac&sVersion=1&webdriver=false

r= requests.get(url=url,headers=headers).text
with open(猫眼榜单.html,w,encoding=utf8) as f:
        f.write(r)
#先把html下载下来便于分析
with open(猫眼榜单.html,r,encoding=utf8) as f:
        html = f.read()
#匹配字体文件
woff_url=re.findall(@font-face.*?format.*?url("(.*?)");},html)[0]  
response = requests.get(url = http:+woff_url).content
with open(猫眼榜单.woff,wb) as f:   
#下载
    f.write(response)

2. 把woff文件保存为xml进行分析字体文件，找到规律

font = TTFont(猫眼榜单.woff)

如果借助xml找不到规律的话，打开woff文件（主要找到字体匹配规则

发现woff字体顺序和cmap对应的顺序是对应的

3. 拼凑数据

num = [9,0,7,3,2,4,6,8,1,5]
cmap = font[cmap].getBestCmap()
print(cmap)
dict={
          
   }
for key,value in cmap.items():
    key=hex(key).replace(0x,"&#x")
    dict[key]=value
print(dict)
i=0
for key in dict.keys():
    if key==&#x78:
        continue
    else:
        dict[key]=num[i]
    i+=1
print(dict)

#得到：
{120: x, 57628: uniE11C, 57708: uniE16C, 58724: uniE564, 58770: uniE592, 59076: uniE6C4, 61163: uniEEEB, 61546: uniF06A, 61913: uniF1D9, 61946: uniF1FA, 62555: uniF45B}
{&#x78: x, &#xe11c: uniE11C, &#xe16c: uniE16C, &#xe564: uniE564, &#xe592: uniE592, &#xe6c4: uniE6C4, &#xeeeb: uniEEEB, &#xf06a: uniF06A, &#xf1d9: uniF1D9, &#xf1fa: uniF1FA, &#xf45b: uniF45B}
{&#x78: x, &#xe11c: 4, &#xe16c: 8, &#xe564: 2, &#xe592: 9, &#xe6c4: 1, &#xeeeb: 3, &#xf06a: 0, &#xf1d9: 5, &#xf1fa: 6, &#xf45b: 7}

4. 最后替换真实的数据

for key,value in dict.items():
          html=html.replace(key+;,str(value))
with open(猫眼榜单解密.html,w,encoding=utf-8)as f:
     f.write(html)

这里其实有点取巧，当字体顺序变了，就不适用了，还是要找规则变化