Re: [Tomahawk-devs] Tomahawk -f <variable> passed

[Koconis, D. <dav...@ic...>]

Sandeep

It is possible to pass a shell variable to tomahawk as you are trying
below.  The error message you posted is usually generated by the
operating system when it can't find a file.  Did you check the
following:

1. The tomahawk executable exists, is in the PATH for the user
attempting to run the command, and that user has permission to execute
it?  Some systems do not put the current working directory in the PATH
so you need to use the full path to tomahawk (e.g.,
/usr/local/bin/tomahawk) instead of just tomahawk.  Also, in your
example below tomahawk is misspelled.  Perhaps this is just a typo in
your posting.  One way to verify all these would be to enter "tomahawk
-?".  If you get the tomahawk help message, then you should be all set.

2. Does the file /home/d/test.pcap exist and does the user have
permission to read it?  Does the tomahawk command run if you do not use
the variable, but instead use the path?

If all of that does not work, please post again with more details about
your configuration.

David

-----Original Message-----
From: tom...@li...
[mailto:tom...@li...] On Behalf Of
KURUPPANKANDI, SANDEEP (SANDEEP)
Sent: Tuesday, February 05, 2008 8:39 AM
To: tom...@li...
Subject: [Tomahawk-devs] Tomahawk -f <variable> passed

Hi,
 I'm facing the following issue:
#########################
pcap=$(zenity --file-selection)
echo $pcap   ## returns the path /home/d/test.pcap
tomahwk -l 1 -f $pcap
#########################
The execution fails with reason :
        "No such file or directory"
 
But when I execute the cmd:
 
**************************************
tomahawk -l 1 -f /home/d/test.pcap
***************************************
Execution passes successfully. Can you please suggest me whether a
variable can be passed as an argument for Tomahawk. If so, why is it
failing in my case.
 
Thanks
-Sandeep