I have a bash that should be run in this way:
./script.sh <arg1> <arg2> <arg3>...<argn>
I want to show these args in my bash:
<arg3> <arg4> ... <argn>
So I wrote this bash:
for (( i=1; i<=$#-3; i++ ))
do
echo $((3+i))
done
but it shows me number of args.
How can I put # in order to see my real args?
Thanks
If you want to show arguments starting from arg3, you can simply use
echo "${@:3}" # OR
printf "%s\n" "${@:3}"
If you really want to show argument indices, use
for (( i=3; i < $#; i++)); do
echo $i
done
You can store all arguments in a BASH array and then use them for processing later:
args=( "$@" )
for (( i=2; i<${#args[@]}; i++ ))
do
echo "arg # $((i+1)) :: ${args[$i]}"
done
A minimal solution that displays the desired arguments without the math:
shift 2
for word
do
echo ${word}
done
I prefer @anubhava's solution of storing the arguments in an array, but to make your original code work, you could use eval:
for ((i=1;i<=$#;i++)); do
eval echo "\$$i"
done
-3 that I wrote..thank you, my problem is: echo $((3+i))
echo ${!i} is a much better method of performing indirect parameter expansion in bash than using eval.After your all good answers I found this solution that works well for my thread:
ARG=( $(echo "${@:3}") )
for (( i=1; i<=$#-3; i++ ))
do
echo ${ARG[i]}
done