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Shell Programming and Scripting

printing last argument in shell script

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# 1  
Old 10-03-2006
printing last argument in shell script
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.
# 2  
Old 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.
# 3  
Old 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.

# 4  
Old 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.
# 5  
Old 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.
# 6  
Old 10-03-2006
Code:
echo "$`echo $#`"

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