oops, remove the ,80 since port is not needed. Well, in my case it wasn't
working with port. notice it gives me 404, but this with my domain
>>> att=urllib2.urlopen(site+payload,80).readlines()
Traceback (most recent call last):
File "", line 1, in
File "/usr/local/lib/python2.6/urllib2.py",
I would try:
site="http://www.bput.org/";
payloads="alert('xss')"
attack= urllib2.urlopen(site+payloads,80).readlines()
-Alex Goretoy
http://www.alexgoretoy.com
somebodywhoca...@gmail.com
On Sun, Jan 11, 2009 at 2:49 AM, Steve Holden wrote:
> Paul Rubin wrote:
> > asit writes:
> >> site="ww
Paul Rubin wrote:
> asit writes:
>> site="www.bput.org"
>> payloads="alert('xss')"
>> attack= urllib2.urlopen(site+payloads,80).readlines()
>>
>> according to my best knowledge, the above code is correct.
>> but why it throws exceptio
>
> The code is incorrect. Look at the string ou are sen
asit writes:
> site="www.bput.org"
> payloads="alert('xss')"
> attack= urllib2.urlopen(site+payloads,80).readlines()
>
> according to my best knowledge, the above code is correct.
> but why it throws exceptio
The code is incorrect. Look at the string ou are sending into
urlopen. What on e
On Sat, Jan 10, 2009 at 9:56 AM, asit wrote:
> site="www.bput.org"
> payloads="alert('xss')"
> attack= urllib2.urlopen(site+payloads,80).readlines()
>
> according to my best knowledge, the above code is correct.
> but why it throws exceptio
Because it's not correct. It's trying to load
www.b
asit wrote:
> site="www.bput.org"
> payloads="alert('xss')"
> attack= urllib2.urlopen(site+payloads,80).readlines()
>
> according to my best knowledge, the above code is correct.
> but why it throws exceptio
what exception it throw?
--
Steve Holden+1 571 484 6266 +1 800 494 3119
H
