①已知MD5值与前面部分内容爆破值

确定长度类

脚本类型1

#coding: utf-8

import hashlib
dic = '0123456789'
for i in range(1000):
  for a in dic:
               t =str(a)+'admin'
               md5 = hashlib.md5(t.encode('utf-8')).hexdigest()
	       #print t
               if md5[:32] == 'ae8b63d93b14eadd1adb347c9e26595a':
                   print (t)

类型2

import string
import hashlib
a='TASC?O3RJMV?WDJKX?ZM'
b='e9032???da???08????911513?0???a2'
dic1=string.digits+string.ascii_lowercase+string.ascii_uppercase
for i1 in dic1:
	for i2 in dic1:
		for i3 in dic1:
			bb='TASC'+i1+'O3RJMV'+i2+'WDJKX'+i3+'ZM'
			aa=hashlib.md5(bb.encode('utf-8'))
			bbb=aa.hexdigest()
			if bbb[:5]=='e9032':
				print i1,i2,i3
import hashlib
a=hashlib.md5('TASCJO3RJMVKWDJKXLZM'.encode('utf-8'))
print a.hexdigest()

截断验证码

import hashlib
def md5(s):
    return hashlib.md5(str(s).encode('utf-8')).hexdigest()
def main(s):
    for i in range(1,99999999):
        if md5(i)[0:6]  == str(s):
            print(i)
            exit(0)
if __name__ == '__main__':
    main("xxxx")


import hashlib
 
dic = 'abcdefghijklmnopqrstuvwxyz1234567890ABCDEFGHIJKLMNOPQRSTUVWXYZ'
 
for i in range(999999999):
    h = hashlib.md5(str(i).encode()).hexdigest()[:6]
    if h == '45ef3d':
        print(i)
        break

不确定长度类

# -*- coding:utf-8 -*-
__author__ = 'Administrator'
#from ultrapower.fd
import itertools as its
import md5

#暴力破解
def uncipher(maxlenth,salt,ciphertext_s,str_letter):
  ciphertext_s=ciphertext_s
  salt = salt
  maxlenth=int(maxlenth)
  str_letter=str_letter
  ciphertext=''
  for i in range(1, maxlenth+1):
    # 迭代生成明文(例如abc,repeat=2  结果为（a,a)(a,b)(a,c)(b,b)(b,a)(b,c)(c,c)(c,a)(c,b)
    r = its.product(str_letter, repeat=i)
    for j in r:
      plaintext = "".join(j) #连接成字符串
      plaintext = "%s%s" % (plaintext, salt)  #把盐加到明文的后面 每次生成的最终明文
      #print plaintext   #打印明文
      # 开始解密，方法是，每个明文进来，加密成密文，然后密文与密文做对比
      md501 = md5.new()
      md501.update(plaintext)
      ciphertext = md501.hexdigest()
      # 对比密文确认明文
      if ciphertext == ciphertext_s:  #如果密文一致 退出2层循环
        break
    if ciphertext == ciphertext_s:    #如果密文一致，退出1层循环，打印结果
      print "task finished(plain,cipher)"
      print "%s:%s" % (plaintext, ciphertext) #打印结果
      break

②与时间戳相互结合的考法

import hashlib
import time
import string
t = str(int(time.time()))#时间戳规范成10位,即消去了float类型状况思路
for i in range(1,100000000000000):
    word = str(t)
    tmp = hashlib.md5(word.encode('utf-8')).hexdigest()#进行md5加密的方法
    x=tmp[0:19]#截断值的作用
    y=x+'admin'
    z=hashlib.md5(y.encode('utf-8')).hexdigest()
    if 'ae8b63d93b14eadd1adb347c9e26595a' == z:
       print (x)

③MD5加密快速构造提交包的一种做法–>直接构造后面的时间进行循环遍历

import requests
import re

def getFlag():
    url = 'http://106.75.214.10:5598/login.php'
    header={
   'cookie':'username=a;session=73c6f70a63b154d81ff45e6555c1a93d'
    }

    html=requests.get(url, headers=header).text
    info=re.findall(r'<h1>([\s\S]*)</h1>',html)
    print(header)
    print (info)

for i in range(1000):
    getFlag()

标签：脚本,hashlib,ciphertext,ctf,str,print,import,md5
来源： https://blog.csdn.net/qq_33942040/article/details/109923453