How to call each argument in a while loop

[toad3000]

I want to access each argument so I tried doing ${$counter2}} because that was my first intuition but I keep getting bad substitution error. Does anyone know how to do this?

typeset -i counter=1typeset -i counter2=1
while [ $counter2 -le $# ] 
do
cat ${$counter2} | while read line 
do
echo "$counter. $line"
counter=counter+1
done
counter2=counter2+1
done

[r-senior]

Are you just trying to access command line arguments? Can we assume you want to write this in Bash shell script?

test.sh

#!/bin/bash

for arg in "$@"
do
    echo $arg
done

$ ./test.sh a b c
a
b
c

[ofnuts]

You don't really need this. If you want to iterate on your arguments, you can use:

for arg in "$@"
do
    echo "Some arg: $arg"
done

The quotes around "$@" are important...

[sisco311]

If the 'in words...' part is ommited then in "$@" is assumed. So you can simply run:

for arg
do
    ...
done

[toad3000]

is it possible to accomplish this using a while loop?

[ofnuts]

If the 'in words...' part is ommited then in "$@" is assumed. So you can simply run:

for arg
do
    ...
done

nice_bash_tricks_I_know++;

[sisco311]

nice_bash_tricks_I_know++;

It's standard. Here are the POSIX specs: http://pubs.opengroup.org/onlinepubs.../contents.html

[sisco311]

is it possible to accomplish this using a while loop?

Yes. But why?

[toad3000]

because I tried many ways to accomplish this using a while loop and It wasn't working. I made a counter and I tried ${$counter} but I get bad substitutional error.

[ofnuts]

It's standard. Here are the POSIX specs: http://pubs.opengroup.org/onlinepubs.../contents.html

But I assume so of course... a "trick" is standard. A "hack" is not.