printing last argument in shell script

arunkumar_mca · #1 (**permalink**) 10-03-2006

All,

I am having a shell script and i will pass different argument diferent time . Please tell me how can i find the last argument that i passsed each time when i exec the script.

Thanks,
Arun.

srikanthus2002 · #2 (**permalink**) 10-03-2006

I think, you have to redirect the value of argument to file
and then echo(by command cat) it where do you want.

vgersh99 · #3 (**permalink**) 10-03-2006

Quote:

Originally Posted by srikanthus2002

I think, you have to redirect the value of argument to file
and then echo(by command cat) it where do you want.

why?

'man ksh' yields the following:

Code:

  Parameters Set by Shell
     The following parameters are automatically set by the shell:

     #         The number of positional parameters in decimal.

aigles · #4 (**permalink**) 10-03-2006

You can do something like this :

Code:

echo "Argument count = $#"
if [ $# -gt 0 ]
then
   eval last_arg=\$$#
else
   last_arg='<none>'
fi
echo "Last argument  = $last_arg"

Jean-Pierre.

petebear · #5 (**permalink**) 10-03-2006

With bash version 3.0 or higher you can use

Code:

${BASH_ARGV[0]}

This little explanation is of courtesy Dark_Helmet over at linxquestions.org

Quote:

BASH_ARGV is a special built-in variable. It's an array of the command line arguments. More specifically, the man page says it's the arguments located on the stack. So it's safe to read them, but I strongly suggest not modifying them. To access the array, use [X] to refer to a specific argument. The arguments are put on the stack "backwards" and explains why the last argument is indexed with 0. Lastly, BASH_ARGV will change inside a function call (because the stack changes). So be careful where/how you access the variable.

bhargav · #6 (**permalink**) 10-03-2006

Code:

echo "$`echo $#`"

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