各種加密邏輯
在進行js逆向的時候，總會遇見一些人類無法直接能理解的東西出現。此時你看到的大多數，是被加密過的密文。
一. 一切從MD5開始
MD5是一個非常常見的摘要(hash)邏輯。其特點就是小巧、速度快、極難被破解(王小云女士)
所以md5，依然是國內非常多的網際網路公司，選擇的密碼摘要演算法
請求物件:
url, method, header, body(page=2)
scrapy 排程器. 為什麼不直接儲存請求物件.
特點：
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這玩意不可逆。 所以摘要演算法，就不是一個加密邏輯.
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相同的內容，計算出來的摘要應該是一樣的
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不同的內容(哪怕是一丟丟丟丟丟不一樣) ，計算出來的結果差別非常大
在數學上. 摘要其實計算邏輯就是hash.
hash(資料) => 數字
1. 密碼
2. 一致性檢測
md5的python實現:
from hashlib import md5
obj = md5()
obj.update("alex".encode("utf-8"))
# obj.update("wusir".encode('utf-8')) # 可以新增多個被加密的內容
bs = obj.hexdigest()
print(bs)
把密文丟到網頁裡，發現有些網站可以直接解密。但其實不然，這裡並不是直接解密MD5，而是"撞庫"
就是它網站裡儲存了大量的MD5的值. 就像這樣:
而需要進行查詢的時候，只需要一條select語句就可以查詢到了，這就是傳說中的撞庫。
如何避免撞庫: md5在進行計算的時候可以加鹽。加鹽之後， 就很難撞庫了
from hashlib import md5
salt = "我是鹽.把我加進去就沒人能破解了"
obj = md5(salt.encode("utf-8")) # 加鹽
obj.update("alex".encode("utf-8"))
bs = obj.hexdigest()
print(bs)
擴充套件：sha256
from hashlib import sha1, sha256
sha = sha256(b'salt')
sha.update(b'alex')
print(sha.hexdigest())
不論是sha1, sha256, md5都屬於摘要演算法. 都是在計算hash值. 只是雜湊的程度不同而已.
這種演算法有一個特性：它們是雜湊，不是加密。而且由於hash演算法是不可逆的，所以不存在解密的邏輯
二. URLEncode和Base64
在訪問一個url的時候，總能看到這樣的一種url
https://www.sogou.com/web?query=%E5%90%83%E9%A5%AD%E7%9D%A1%E8%A7%89%E6%89%93%E8%B1%86%E8%B1%86&_asf=www.sogou.com&_ast=&w=01019900&p=40040100&ie=utf8&from=index-nologin&s_from=index&sut=3119&sst0=1630994614300&lkt=0%2C0%2C0&sugsuv=1606978591882752&sugtime=1630994614300
此時會發現, 在瀏覽器上明明是能看到中文的. 但是一旦複製出來. 或者在抓包工具裡看到的. 都是這種%. 那麼這個%是什麼鬼? 也是加密麼?
非也，其實在訪問一個url的時候，瀏覽器會自動的進行urlencode操作，會對請求的url進行編碼
這種編碼規則被稱為百分號編碼，是專門為url(統一資源定位符)準備的一套編碼規則.
一個url的完整組成:
scheme://host:port/dir/file?p1=v1&p2=v2#anchor
http ://www.baidu.com/tieba/index.html?name=alex&age=18
引數: key=value 的形式 # 伺服器可以透過key拿value
此時，如果引數中出現一些特殊符號，比如'=' ，想給伺服器傳遞 ‘a=b=c’ 這樣的引數，必然會讓整個URL產生歧義
所以,把url中的引數部分轉化成位元組，每位元組的再轉化成2個16進位制的數字，前面補%.
看著很複雜，在python裡，直接一步到位
from urllib.parse import urlencode, unquote, quote
# 單獨編碼字串 quote unqquote 解碼
wq = "米飯怎麼吃"
print(quote(wq)) # %E7%B1%B3%E9%A5%AD%E6%80%8E%E4%B9%88%E5%90%83
print(quote(wq, encoding="gbk")) # %C3%D7%B7%B9%D4%F5%C3%B4%B3%D4
# 多個資料統一進行編碼 urlencode
dic = {
"wq": "米飯怎麼吃",
"new_wq": "想怎麼吃就怎麼吃"
}
print(urlencode(dic)) # wq=%E7%B1%B3%E9%A5%AD%E6%80%8E%E4%B9%88%E5%90%83&new_wq=%E6%83%B3%E6%80%8E%E4%B9%88%E5%90%83%E5%B0%B1%E6%80%8E%E4%B9%88%E5%90%83
print(urlencode(dic, encoding="utf-8")) # 也可以指定字符集
# 一個完整的url編碼過程
base_url = "http://www.baidu.com/s?"
params = {
"wd": "大王"
}
url = base_url + urlencode(params)
print(url) # http://www.baidu.com/s?wd=%E5%A4%A7%E7%8E%8B
解碼
s = "http://www.baidu.com/s?wd=%E5%A4%A7%E7%8E%8B"
print(unquote(s)) # http://www.baidu.com/s?wd=大王
base64
通常被加密後的內容是位元組，而密文是用來傳輸的（不傳輸誰加密啊）
但在http協議裡，想要傳輸位元組是很麻煩的一個事兒。相對應的， 如果傳遞的是字串就好控制的多，此時base64就應運而生了。
26個大寫字母 + 26個小寫字母 + 10個數字 + 2個特殊符號( + / )，組成了一組類似64進位制的計算邏輯，這就是base64了。
import base64
bs = "我要吃飯".encode("utf-8")
# 把位元組轉化成b64
print(base64.b64encode(bs).decode())
# 把b64字串轉化成位元組
s = "5oiR6KaB5ZCD6aWt"
print(base64.b64decode(s).decode("utf-8"))
注意：
b64處理後的字串長度， 一定是4的倍數。如果在網頁上，看到有些密文的b64長度，不是4的倍數，會報錯
eg:
import base64
s = "ztKwrsTj0b0"
bb = base64.b64decode(s)
print(bb)
# 此時執行出現以下問題
Traceback (most recent call last):
File "D:/PycharmProjects/rrrr.py", line 33, in <module>
bb = base64.b64decode(s)
File "D:\Python38\lib\base64.py", line 87, in b64decode
return binascii.a2b_base64(s)
binascii.Error: Incorrect padding
解決思路：base64長度要求，字串長度必須是4的倍數， 用 = 填充一下即可
import base64
s = "ztKwrsTj0b0"
s += ("=" * (4 - len(s) % 4))
print("填充後", s)
bb = base64.b64decode(s).decode("gbk")
print(bb)
三. 對稱加密
所謂對稱加密，就是加密和解密用的是同一個秘鑰
就好比：我要給你郵寄一個箱子，上面懟上鎖，提前我把鑰匙給了你一把、我一把。
那麼我在郵寄之前，就可以把箱子鎖上，然後快遞到你那裡，你用相同的鑰匙，就可以開啟這個箱子
條件: 加密和解密用的是同一個秘鑰。那麼兩邊就必須同時擁有鑰匙，才可以
常見的對稱加密: AES、DES、3DES。 這裡討論AES、DES
3.1 AES
# AES加密
from Crypto.Cipher import AES
"""
長度
16: *AES-128*
24: *AES-192*
32: *AES-256*
引數：MODE 加密模式. 常見的ECB, CBC
以下內容來自網際網路~~
ECB：是一種基礎的加密方式，密文被分割成分組長度相等的塊（不足補齊），然後單獨一個個加密，一個個輸出組成密文。
CBC：是一種迴圈模式，前一個分組的密文和當前分組的明文異或或操作後再加密，這樣做的目的是增強破解難度。
CFB/OFB：實際上是一種反饋模式，目的也是增強破解的難度。
FCB和CBC的加密結果是不一樣的，兩者的模式不同，而且CBC會在第一個密碼塊運算時加入一個初始化向量。
"""
aes = AES.new(b"alexissbalexissb", mode=AES.MODE_CBC, IV=b"0102030405060708")
data = "我吃飯了"
data_bs = data.encode("utf-8")
# 需要加密的資料必須是16的倍數
# 填充規則: 缺少資料量的個數 * chr(缺少資料量個數)
pad_len = 16 - len(data_bs) % 16
data_bs += (pad_len * chr(pad_len)).encode("utf-8")
bs = aes.encrypt(data_bs)
print(bs)
AES解密
from Crypto.Cipher import AES
aes = AES.new(b"alexissbalexissb", mode=AES.MODE_CBC, IV=b"0102030405060708")
# 密文
bs = b'\xf6z\x0f;G\xdcB,\xccl\xf9\x17qS\x93\x0e'
result = aes.decrypt(bs) # 解密
print(result.decode("utf-8"))
3.2 DES
# DES加密解密
from Crypto.Cipher import DES
# key: 8個位元組
des = DES.new(b"alexissb", mode=DES.MODE_CBC, IV=b"01020304")
data = "我要吃飯".encode("utf-8")
# 需要加密的資料必須是16的倍數
# 填充規則: 缺少資料量的個數 * chr(缺少資料量個數)
pad_len = 8 - len(data) % 8
data += (pad_len * chr(pad_len)).encode("utf-8")
bs = des.encrypt(data)
print(bs)
# 解密
des = DES.new(key=b'alexissb', mode=DES.MODE_CBC, IV=b"01020304")
data = b'6HX\xfa\xb2R\xa8\r\xa3\xed\xbd\x00\xdb}\xb0\xb9'
result = des.decrypt(data)
print(result.decode("utf-8"))
四.非對稱加密
非對稱加密，加密和解密的秘鑰不是同一個秘鑰。
需要兩把鑰匙：一個公鑰（公開的秘鑰, 對資料進行加密）、一個私鑰（私密的秘鑰, 對資料進行解密）
公鑰傳送給客戶端 (傳送端)，傳送端用公鑰對資料進行加密，再傳送給接收端，接收端使用私鑰來對資料解密。
由於私鑰只存放在接受端這邊，所以即使資料被截獲了，也是無法進行解密的
# 非對稱加密的邏輯:
1. 先在伺服器端. 生成一組秘鑰, 公鑰/私鑰
2. 把公鑰放出去，傳送給客戶端.
3. 客戶端在拿到公鑰之後. 可以使用公鑰對資料進行加密.
4. 把資料傳輸給伺服器
5. 伺服器使用私鑰，對資料進行解密.
常見的非對稱加密演算法: RSA、DSA等等。介紹一個：RSA加密，也是最常見的一種加密方案
4.1 RSA加密解密
4.1.1 建立公鑰和私鑰
from Crypto.PublicKey import RSA
from Crypto.Cipher import PKCS1_v1_5
from Crypto import Random
import base64
# 隨機
gen_random = Random.new
# 生成秘鑰
rsakey = RSA.generate(1024)
with open("rsa.public.pem", mode="wb") as f:
f.write(rsakey.publickey().exportKey())
with open("rsa.private.pem", mode="wb") as f:
f.write(rsakey.exportKey())
4.1.2 加密
# 加密
data = "我要吃飯了"
with open("rsa.public.pem", mode="r") as f:
pk = f.read()
rsa_pk = RSA.importKey(pk)
rsa = PKCS1_v1_5.new(rsa_pk)
result = rsa.encrypt(data.encode("utf-8"))
# 處理成b64方便傳輸
b64_result = base64.b64encode(result).decode("utf-8")
print(b64_result)
4.1.3 解密
data = "e/spTGg3roda+iqLK4e2bckNMSgXSNosOVLtWN+ArgaIDgYONPIU9i0rIeTj0ywwXnTIPU734EIoKRFQsLmPpJK4Htte+QlcgRFbuj/hCW1uWiB3mCbyU3ZHKo/Y9UjYMuMfk+H6m8OWHtr+tWjiinMNURQpxbsTiT/1cfifWo4="
# 解密
with open("rsa.private.pem", mode="r") as f:
prikey = f.read()
rsa_pk = RSA.importKey(prikey)
rsa = PKCS1_v1_5.new(rsa_pk)
result = rsa.decrypt(base64.b64decode(data), gen_random)
print(result.decode("utf-8"))
五. 案例
5.1 藝恩電影資料解密
以https://www.endata.com.cn/BoxOffice/BO/Year/index.html為案例. 來完成該網站資料解密
分析: 很明顯，該網站的資料是經過加密的
接下來，到 Initiator （啟動器，請求呼叫堆疊 看得到傳送請求的 依次觸發的檔案） 裡看看
接下來就是逆向的過程了... 各位..還是看影片吧. 影片裡有詳細的分析過程.
給出完整破解程式碼
import binascii # 二進位制和ascii之間轉換
from Crypto.Cipher import DES
def func(a, b, c):
if b == 0:
return a[c:]
d = a[:b] + a[b+c:]
return d
def process(data):
e = int(data[len(data)-1], base=16) + 9
f = int(data[e], base=16)
data = func(data, e, 1)
e = data[f:f+8]
data = func(data, f, 8)
jiemi(data, str(e), str(e))
def jiemi(data, key, iv):
des = DES.new(key.encode("utf-8"), mode=DES.MODE_ECB)
# de_text = base64.standard_b64decode(data)
# 十六進位制表示的二進位制資料 -> 十六進位制 -> 二進位制
print(binascii.a2b_hex(data))
ee = des.decrypt(binascii.a2b_hex(data))
print(ee.decode("utf-8"))
if __name__ == '__main__':
data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process(data)
5.2 中大網校登入
接下來開始逆向試試吧. 從登入位置開始，第一個事兒要搞定的，是那個煩人的驗證碼
但是這個請求是需要cookie的，想想也應該如此。因為伺服器要知道這張圖片給哪個客戶端使用了。就必須藉助cookie，來判別不同的客戶端.
所以. 整個流程應該是:
- 進入登入頁，載入到cookie
- 訪問驗證碼ur，獲取到驗證碼，並完成破解
- 訪問getTime api，雖然不知道它用來做什麼， 但是在後續的密碼加密時，是需要這個api返回的data
- 準備好使用者名稱和密碼，對密碼進行加密
- 傳送登入請求
- 得到的結果處理(加入到cookie中)
程式碼:
import requests
import json
import time
from Crypto.Cipher import PKCS1_v1_5
from Crypto.PublicKey import RSA
import base64
def base64_api(img, uname='q6035945', pwd='q6035945', typeid=3):
data = {"username": uname, "password": pwd, "typeid": typeid, "image": img}
result = json.loads(requests.post("http://api.ttshitu.com/predict", json=data).text)
if result['success']:
return result["data"]["result"]
else:
return result["message"]
def enc(s):
key = "MIGfMA0GCSqGSIb3DQEBAQUAA4GNADCBiQKBgQDA5Zq6ZdH/RMSvC8WKhp5gj6Ue4Lqjo0Q2PnyGbSkTlYku0HtVzbh3S9F9oHbxeO55E8tEEQ5wj/+52VMLavcuwkDypG66N6c1z0Fo2HgxV3e0tqt1wyNtmbwg7ruIYmFM+dErIpTiLRDvOy+0vgPcBVDfSUHwUSgUtIkyC47UNQIDAQAB"
rsa_key = RSA.importKey(base64.b64decode(key))
rsa_new = PKCS1_v1_5.new(rsa_key)
mi = rsa_new.encrypt(s.encode("utf-8"))
return base64.b64encode(mi).decode("utf-8")
session = requests.session()
session.headers = {
'Referer' : "https://user.wangxiao.cn/login?url=http%3A%2F%2Fks.wangxiao.cn%2F",
"Content-Type": "application/json;charset=UTF-8",
"User-Agent": "Mozilla/5.0 (Macintosh; Intel Mac OS X 10_15_4) AppleWebKit/537.36 (KHTML, like Gecko) Chrome/96.0.4664.110 Safari/537.36"
}
login_url = "https://user.wangxiao.cn/login?url=http%3A%2F%2Fks.wangxiao.cn%2F"
# 進入登入頁面(目的: 載入到第一波cookie)
session.get(login_url)
time.sleep(1)
resp = session.post("https://user.wangxiao.cn/apis//common/getImageCaptcha")
dic = resp.json()
operation_date = dic['operation_date']
img_data = dic['data'].split(",")[1]
verify_code = base64_api(img_data)
# 走登入流程
# 1.getTime
get_time_resp = session.post("https://user.wangxiao.cn/apis/common/getTime")
get_time_dic = get_time_resp.json()
get_time = get_time_dic['data']
# 2. 登入
login_verify_url = "https://user.wangxiao.cn/apis/login/passwordLogin"
username = "你的賬號"
password = "你的密碼"
data = {
"imageCaptchaCode": verify_code,
"password": enc(password+get_time),
"userName": username,
}
resp = session.post(login_verify_url, data=json.dumps(data))
dic = resp.json()['data']
"""
window.syncLogin = function(e, o) {
// util.js
var n = {
path: "/",
domain: "wangxiao.cn"
};
o && (n.expires = o),
$.cookie("UserCookieName", e.userName, n),
$.cookie("OldUsername2", e.userNameCookies, n),
$.cookie("OldUsername", e.userNameCookies, n),
$.cookie("OldPassword", e.passwordCookies, n),
$.cookie("UserCookieName_", e.userName, n),
$.cookie("OldUsername2_", e.userNameCookies, n),
$.cookie("OldUsername_", e.userNameCookies, n),
$.cookie("OldPassword_", e.passwordCookies, n),
e.sign && (n.expires = 365,
$.cookie(e.userName + "_exam", e.sign, n))
//index.js
keepOurCookie12("autoLogin",null) ;
keepOurCookie12("userInfo",JSON.stringify(res.data)) ;
keepOurCookie12("token",res.data.token) ;
}
"""
session.cookies['UserCookieName'] = dic['userName']
session.cookies['OldUsername2'] = dic['userNameCookies']
session.cookies['OldUsername'] = dic['userNameCookies']
session.cookies['OldPassword'] = dic['passwordCookies']
session.cookies['UserCookieName_'] = dic['userName']
session.cookies['OldUsername2_'] = dic['userNameCookies']
session.cookies['OldUsername_'] = dic['userNameCookies']
session.cookies['OldPassword_'] = dic['passwordCookies']
session.cookies['autoLogin'] = "null"
session.cookies['userInfo'] = json.dumps(dic)
session.cookies['token'] = dic['token']
data = {
"examPointType": "",
"practiceType": "2",
"questionType": "7",
"sign": "jz1",
"subsign": "8cc80ffb9a4a5c114953",
"top": "30"
}
resp = session.post('http://ks.wangxiao.cn/practice/listQuestions', data=json.dumps(data))
print(resp.text)