Using i as a variable in bash shell script

Question

I am trying to create a shell script that will take command line arguments and create one or more files based off of that. I know that each command line argument is stored in $0, $1, $2...and so on, so that is what this loop is based on.

for i in $(eval echo {1..$#})
do
    echo "File I'm about to edit/create: "$i""
    touch "$i"
done

However, $i is being taken literally as the number 1, 2..rather than the value in $1.

Thanks! That was actually really helpful. I can do this whole routine without even using a for loop. — user2793442
– user2793442, Commented Oct 2, 2014 at 20:28

chepner · Accepted Answer · 2014-10-02 20:28:21Z

2

You would need to use indirect parameter expansion:

echo "File...: ${!i}"
touch "${!i}"

However, it's much simpler to just iterate over the arguments themselves:

for f in "$@"; do
    echo "File...: $f"
    touch "$f"
done

answered Oct 2, 2014 at 20:28

chepner

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