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破解字体加密解决思路

 之前做过爬取大众点评美团等电话号码遇到过字体反爬  ,故此处做个人记录

 例如:58同城、起点中文网、大众点评等

加密思路

原理:关于字体加密,其实是将一种特定的字体库来代替浏览器本身的字体库显示的过程

以58字体库加密为例

58同城中,无论是简历中的字体加密,还是房产信息中的加密都是有迹可循的;正如我们所知,加密的字体数量一般都不会太大且会随访问请求而变化,也就是说 ,字体加密的字体库每次都是变化的,我们需要在爬去网页的同时拿取加密的字体库,进行解密,

基本流程:

1、请求页面

2、获取加密的字体库

3、解析字体库,获取字体间的映射关系

4、获取加密的字体,获取字体间映射关系,一一对应

字体加密一般是网页修改了默认的字符编码集,在网页上加载的他们自己定义的字体文件作为字体的样式,可以正确地显示数字,但是在源码上同样的二进制数由于未加载自定义的字体文件就由计算机默认编码成了乱码。

网站url:https://zz.58.com/pinpaigongyu/?utm_source=sem-baidu-pc&spm=105916146708.26420796287&PGTID=0d100000-0015-63fb-169b-74ddef51914e&ClickID=2

破解字体加密解决思路

一般来说,通用的解决办法是找到字体文件,分析文件中的映射关系。一般来说,字体文件都是作为样式加在加密字体的部位。

在样式中,我基本上都看完了, 从名字上看只有这个

fangchan-secret

最可能是字体加密文件。

破解字体加密解决思路

在源码中

Ctrl+F

搜索

fangchan-secret

 寻找字体加密文件

破解字体加密解决思路

在58的源码中,字体文件是通过base64加密之后放在js里面了。把其中加密的部分取出,第一次是分析,在代码中可使用正则将其中的内容取出来。

58的字体加密文件每次网页刷新,其中的映射顺序会变,所以在不刷新的情况下,再复制一份

这里取第一条799 及其对应的乱码。

base64_str = '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'
price_code = '鸺齤齤'
           

首先进行base64解码,转化成为二进制形式,在方法中同时我也将字体文件写入了otf字体文件中。

def make_font_file(base64_string: str):
    bin_data = base64.decodebytes(base64_string.encode())
    with open('text.otf','wb') as f:
        f.write(bin_data)
    return bin_data
           

使用fontCreator打开字体文件可以直观的看到每一个数字对应的编码

破解字体加密解决思路

将字节文件转化为xml格式

def convert_font_to_xml(bin_data):
    # 由于TTFont接收一个文件类型
    # BytesIO(bin_data) 把二进制数据当作文件来操作
    font = TTFont(BytesIO(bin_data))
    font.saveXML("text.xml")
bin_data = make_font_file(base64_str)
convert_font_to_xml(bin_data)
# 获取对应关系
font = TTFont(BytesIO(make_font_file(base64_str)))
uniList = font['cmap'].tables[0].ttFont.getGlyphOrder()
c = font['cmap'].tables[0].ttFont.tables['cmap'].tables[0].cmap
# c = font.getBestCmap()
print('cmap is:::::', c)
           

打印出来是这样子:cmap is::::: {38006: 'glyph00007', 38287: 'glyph00009', 39228: 'glyph00004', 39499: 'glyph00005', 40506: 'glyph00008', 40611: 'glyph00006', 40804: 'glyph00010', 40850: 'glyph00001', 40868: 'glyph00003', 40869: 'glyph00002'}

该字典的键就是网页上显示的乱码的unicode编码,值就是该乱码对应的真正数字。

其中glyph00007是资源,每一个glyph0000x对应一个数字 、58的字体文件比较偷懒,根据其后缀就知道对应的数字是对少

又字体的xml文件中,下图部分开始可知(我没截取完),glyph00000没有意义,glyph00001对应0,glyph00001对应1以此类推。

破解字体加密解决思路

那么,我们根据网页抓取的乱码的unicode编码,获取其对对应的字源,即可获取所对应的数字。

def get_num(string):
    ret_list = []
    for char in string:
        decode_num = ord(char)
        num = c[decode_num]
        num = int(num[-2:])-1
        ret_list.append(num)
    return ret_list
           

如果通过浏览器看到的是类似

鸺齤齤

这样的乱码,使用爬虫获取的数据是类似 

鸺龒龒

 十六进制的数字,可直接截取后面四位转化为十进制数后在通过映射表查找。

以下是详细代码

import base64
from io import BytesIO
from fontTools.ttLib import TTFont
import requests
import re
from lxml import etree
 
 
url = 'https://cd.58.com/wuhou/chuzu/b5j5'
res = requests.get(url)
bs64_str = re.findall("charset=utf-8;base64,(.*?)'\)", res.text)[0]
 
 
def get_page_show_ret(string):
    font = TTFont(BytesIO(base64.decodebytes(bs64_str.encode())))
    c = font['cmap'].tables[0].ttFont.tables['cmap'].tables[0].cmap
    ret_list = []
    for char in string:
        decode_num = ord(char)
        if decode_num in c:
            num = c[decode_num]
            num = int(num[-2:])-1
            ret_list.append(num)
        else:
            ret_list.append(char)
    ret_str_show = ''
    for num in ret_list:
        ret_str_show += str(num)
    return ret_str_show
 
 
page = etree.HTML(res.text)
li = page.xpath('.//ul[@class="listUl"]//li')[0:-1]
for each_li in li:
    title = each_li.xpath('.//div[@class="des"]/h2/a/text()')[0].strip()
    title = get_page_show_ret(title)
    price = each_li.xpath('.//div[@class="money"]/b/text()')[0]
    price = get_page_show_ret(price)
    print(title)
    print(price)
    print('='*20)
           

继续阅读