有没有做电子名片的网站地推拉新接单网

码上爬-第7&8题

靶场地址:

7: 码上爬 - 全网一站式爬虫刷题平台

8: 码上爬 - 全网一站式爬虫刷题平台

js加密 & js混淆

题七：千山鸟飞绝

打开控制台有无限debugger

找了个油猴插件hook掉: remove-debugger

查看接口调用堆栈的 pagenation.js文件, 已经被混淆

全部复制找个解混淆的网站尝试还原一下

解混淆测试版

js反混淆 - dejs.vip

还原后替换原有混淆的js文件

可以看到, param参数的x为请求头的参数m+固定值"xxoo" , 经过SHA256算法得到

验证SHA256算法: SHA256 - 在线工具

也就是只要确定了方法 window["eeee"] 的算法后, 参数m,x都出来了

正好看到AES的解密, 猜测应该是响应的解密, 验证一下; 证明确实对的, 这样解密算法也出来了, 就只剩eeee方法的实现了

由于eeee方法还调用了很多其他方法和变量, 需要一个一个补; 为了省事, 直接把整个解混淆后的js拿过来修改 尝试直接调用eeee方法

取消 IIFE（Immediately Invoked Function Expression，立即调用函数表达式）的独立作用域, 可以直接在外部调用, 或者在独立作用域内调用

保证入参一致, 验证返回值是否相同; 验证值一致, 就可以直接python调用js方法

还有给AI js代码, 让其使用python实现, Ai会知道eeee方法其实就是个MD5, 验证一下

至此, 所有加解密算法已经出来了, 调个接口试试

使用的加密库: Examples — PyCryptodome 3.230b0 documentation

pip install pycryptodomex

最后做个翻页请求20次接口累加数组即可得到答案, 就不做了, 贴个代码

import json
import time
import requests
import hashlib
from Cryptodome.Cipher import AES
from Cryptodome.Util.Padding import unpadcookies = {"sessionid": "66cxjeayq4hhd6e0bedtqze62lluxzhe","Hm_lvt_b5d072258d61ab3cd6a9d485aac7f183": "1743067108,1743124399","HMACCOUNT": "64ECCFF2BFC433A1","Hm_lpvt_b5d072258d61ab3cd6a9d485aac7f183": "1743155397",
}timestamp = str(int(time.time() * 1000))
# 调用 JS 函数
m = hashlib.md5(("xialuo" + timestamp).encode()).hexdigest()
print(m)
x = hashlib.sha256((m + "xxoo").encode()).hexdigest()
print(x)
headers = {"accept": "application/json, text/javascript, */*; q=0.01","accept-language": "zh-CN,zh;q=0.9,en;q=0.8,en-GB;q=0.7,en-US;q=0.6","cache-control": "no-cache","m": m,"pragma": "no-cache","priority": "u=1, i","referer": "https://stu.tulingpyton.cn/problem-detail/7/","sec-ch-ua": '"Chromium";v="134", "Not:A-Brand";v="24", "Microsoft Edge";v="134"',"sec-ch-ua-mobile": "?0","sec-ch-ua-platform": '"Windows"',"sec-fetch-dest": "empty","sec-fetch-mode": "cors","sec-fetch-site": "same-origin","ts": timestamp,"user-agent": "Mozilla/5.0 (Windows NT 10.0; Win64; x64) AppleWebKit/537.36 (KHTML, like Gecko) Chrome/134.0.0.0 Safari/537.36 Edg/134.0.0.0","x-requested-with": "XMLHttpRequest",# 'cookie': 'sessionid=66cxjeayq4hhd6e0bedtqze62lluxzhe; Hm_lvt_b5d072258d61ab3cd6a9d485aac7f183=1743067108,1743124399; HMACCOUNT=64ECCFF2BFC433A1; Hm_lpvt_b5d072258d61ab3cd6a9d485aac7f183=1743155397',
}params = {"page": "1","x": x,
}response = requests.get("https://stu.tulingpyton.cn/api/problem-detail/7/data/",params=params,cookies=cookies,headers=headers,
)
json_data = response.json()
print(json_data)
r = json_data['r']
# hex to bytes
r = bytes.fromhex(r)key = b"xxxxxxxxoooooooo"
iv = b"0123456789ABCDEF"
chiper = AES.new(key=key, mode=AES.MODE_CBC, iv=iv)
result = unpad(chiper.decrypt(r), AES.block_size)
print(result)
result = json.loads(result)
print(result)
print(result['current_array'])

题八：迷踪步

依旧还是解混淆替换, 然后跟栈断点 定位加密位置

尝试分析加密方法OOOoOo, 先分析参数, 经过几次请求断点, 基本确定"ooooo"为固定值, 只有一个时间戳变量, 但是加密逻辑实在有点难看懂, 丢给AI试试

拿Ai生成的方法测试返回值是否一致, 没想到还真出来了; 甚至优化后, 加密逻辑可简化至两行代码,

贴个代码; 以学习为目的的话, 还是应该尝试慢慢扣代码还原逻辑, 而不是丢给AI🫡

import base64
import requestsimport timedef encrypt_string(timestamp):key_string = "oooooo"input_string = key_string + timestamp + "1"input_chars = list(input_string)key_chars = list(key_string)result = []# Process input in chunks of 4 charactersfor i in range(0, len(input_chars), 4):chunk = input_chars[i:i+4]# XOR each character in the chunk with corresponding key characterfor j in range(len(chunk)):input_char_code = ord(chunk[j])key_char_code = ord(key_chars[j % len(key_chars)])encrypted_code = (input_char_code + key_char_code) % 256chunk[j] = chr(encrypted_code)result.extend(chunk)# Convert to hexadecimal representationencrypted_string = ''.join(result)hex_string = ''.join(f"{ord(char):02x}" for char in encrypted_string)return hex_stringdef encrypt_string1(timestamp):input_str = "oooooo" + timestamp + "1"return ''.join(f"{(ord(c) + 111) % 256:02x}" for c in input_str)cookies = {'Hm_lvt_b5d072258d61ab3cd6a9d485aac7f183': '1743067108','HMACCOUNT': '64ECCFF2BFC433A1','sessionid': '66cxjeayq4hhd6e0bedtqze62lluxzhe','Hm_lpvt_b5d072258d61ab3cd6a9d485aac7f183': '1743072264','s': '51b351b351b351b370b0d03050b010d07150b07130',
}timestamp = str(int(time.time() * 1000))m = encrypt_string1(timestamp)
t = base64.b64encode(timestamp.encode()).decode()print(m, t)
headers = {'accept': '*/*','accept-language': 'zh-CN,zh;q=0.9,en;q=0.8,en-GB;q=0.7,en-US;q=0.6','cache-control': 'no-cache','content-type': 'application/json','m': m,'origin': 'https://stu.tulingpyton.cn','pragma': 'no-cache','priority': 'u=1, i','referer': 'https://stu.tulingpyton.cn/problem-detail/8/','sec-ch-ua': '"Chromium";v="134", "Not:A-Brand";v="24", "Microsoft Edge";v="134"','sec-ch-ua-mobile': '?0','sec-ch-ua-platform': '"Windows"','sec-fetch-dest': 'empty','sec-fetch-mode': 'cors','sec-fetch-site': 'same-origin','t': t,'user-agent': 'Mozilla/5.0 (Windows NT 10.0; Win64; x64) AppleWebKit/537.36 (KHTML, like Gecko) Chrome/134.0.0.0 Safari/537.36 Edg/134.0.0.0','x-requested-with': 'XMLHttpRequest',# 'cookie': 'Hm_lvt_b5d072258d61ab3cd6a9d485aac7f183=1743067108; HMACCOUNT=64ECCFF2BFC433A1; sessionid=66cxjeayq4hhd6e0bedtqze62lluxzhe; Hm_lpvt_b5d072258d61ab3cd6a9d485aac7f183=1743072264; s=51b351b351b351b370b0d03050b010d07150b07130',
}json_data = {'page': 1,
}response = requests.post('https://stu.tulingpyton.cn/api/problem-detail/8/data/', cookies=cookies, headers=headers, json=json_data)print(response.json())

51b351b351b351b370b0d03050b010d07150b07130’,
}

json_data = {
‘page’: 1,
}

response = requests.post(‘https://stu.tulingpyton.cn/api/problem-detail/8/data/’, cookies=cookies, headers=headers, json=json_data)

print(response.json())