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/9699919799/utilities/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/9699919799/utilities/contents.html
But I assume so of course... a "trick" is standard. A "hack" is not.

[Vaphell]

i=1; while (( i<=$# )); do echo "\$$i: ${!i}"; ((i++)); done


$ ./arg a bb ccc dddd
$1: a
$2: bb
$3: ccc
$4: dddd

or

while (( $# )); do echo "$1"; shift; done;
this one is destructive (consumes params)

[sisco311]

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.

Oh, I see. You are trying to use indirect expansion. As you can see from the example posted by Vaphell, the format of indirect expansion is, ${!name}:


name=1
echo "\$$name is: ${!name}"


For more details please check out BashFAQ 006 (link in my signature) and the `Parameter Expansion' section from the man page of bash:

man bash | less '+/^ +Parameter Expansion'