2023年7月31日发(作者：)

利⽤Python暴⼒爆破PDF密码⼀个简单的Python脚本，可⽤于暴⼒破解受密码保护的PDF⽂件的密码脚本已在使⽤128位RC4（⼤多数信⽤卡对帐单）加密的PDF上进⾏了测试，成功率为100％pasword=''--decryptImport sysfrom pyPdf import PdfFileReaderhelpmsg = "Simple PDF brute force scriptn"helpmsg += "Cracks pwds of the format 0000-9999."helpmsg += "Example: snow0653nn"helpmsg += "Usage: "if len() < 2: print helpmsg ()

pdffile = PdfFileReader(file([1], "rb"))if ypted == False: print "[!] The file is not protected with any password. Exiting." exitprint "[+] Attempting to Brute force. This could take "z = ""for i in range(0,9999): z = str (i) while (len(z) < 4): z = "0" + z

a = str([2][:4] + str(z))

if t(a) > 0: print "[+] Password is: " + a print "[...] Exiting.." ()

2023年7月31日发(作者：)

利⽤Python暴⼒爆破PDF密码⼀个简单的Python脚本，可⽤于暴⼒破解受密码保护的PDF⽂件的密码脚本已在使⽤128位RC4（⼤多数信⽤卡对帐单）加密的PDF上进⾏了测试，成功率为100％pasword=''--decryptImport sysfrom pyPdf import PdfFileReaderhelpmsg = "Simple PDF brute force scriptn"helpmsg += "Cracks pwds of the format 0000-9999."helpmsg += "Example: snow0653nn"helpmsg += "Usage: "if len() < 2: print helpmsg ()

pdffile = PdfFileReader(file([1], "rb"))if ypted == False: print "[!] The file is not protected with any password. Exiting." exitprint "[+] Attempting to Brute force. This could take "z = ""for i in range(0,9999): z = str (i) while (len(z) < 4): z = "0" + z

a = str([2][:4] + str(z))

if t(a) > 0: print "[+] Password is: " + a print "[...] Exiting.." ()