How to use param values inside for loop I am running my script as `script.sh 12345 12346 12347` for z in 1..$(seq 1 $#); do echo "param $z is $($(echo $z))"; //Line 4 done; I am expecting output as below: param 1 is 12345 param 2 is 12346 param 3 is 12347 Guess, I am missing something in Line 4.

The problem with `$($(echo $z))` is that it expands, first to `$(1)` (if `$z` is 1) and then the shell tries to run `1` as a command.

* * *

Assuming `bash`:


params=( "$@" )

for (( i = 0; i < ${#params[@]}; ++i )); do
printf 'Param %d is "%s"\
' "$i" "${params[i]}"
done


Running it:


$ bash script.sh a b c
Param 0 is "a"
Param 1 is "b"
Param 2 is "c"


Or, with `/bin/sh`:


i=0
while [ "$#" -gt 0 ]; do
printf 'Param %d is "%s"\
' "$i" "$1"
i=$(( i + 1 ))
shift
done


Running that:


$ /bin/sh script.sh a b c
Param 0 is "a"
Param 1 is "b"
Param 2 is "c"


That's if you really need to explicitly enumerate them. Usually one would just loop over `"$@"`:


for param in "$@"; do
printf 'Param: "%s"\
' "$param"
# do other thing with "$param" here
done