I created the following script:

*#!/bin/bash
QUERY_STRING=$1
cat /tmp/`echo $QUERY_STRING | sed "s/-//g"`.ext*

Then I ran this from the command line:

*sh test.sh `echo "blah">test.foo`
*
The file 'test.foo' was created as expected. I changed the script to look
like this and it also wrote the file:

*#!/bin/bash
QUERY_STRING="`echo "blah">test.foo`"
cat /tmp/`echo $QUERY_STRING | sed "s/-//g"`.ext*


On Tue, Jun 1, 2010 at 5:03 PM, Robin Wood <[email protected]> wrote:

> Earlier today I was asked if I could work out a way to get command
> injection through the code below. I couldn't work out a way to do it
> but hopefully (here's looking at you Ed) someone else can work
> something out.
>
> The vulnerable line is:
>
> cat /tmp/`echo $QUERY_STRING | sed "s/-//g"`.ext
>
> and I'm looking for command injection or a way to create a new file as
> this is on a web server so I could create a web page shell script. I
> know that I've got the ability to read any file, I'm not after that.
>
> The obvious thing to do is to set $QUERY_STRING to something like
>
> x ; echo 'hi' > /www/hi.html
>
> but that doesn't work. I get
>
> cat: x: No such file or directory
> cat: ;: No such file or directory
> cat: echo: No such file or directory
> cat: 'hi': No such file or directory
> cat: >: No such file or directory
> cat: /www/hi.html.ext: No such file or directory
>
> however if I put the value of QUERY_STRING directly into the statement
> then it does work and the file is created. The fact that QUERY_STRING
> is a variable means it is treated differently than if the commands
> were just put inline.
>
> Any tips?
>
> Robin
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