Shell Scripting: Generating output file name in script and then writing to it

Question

Hello,

I have a shell script where a user passes different sizes as command line arguments. For each of those sizes, I want to perform some task, then save the output of that task into a .txt file with the size in the name.

1. How can I take the command line passed and make that part of a stringname for the output file? I save the output file in a directory specified by another command line argument. Perhaps an example will clear it up.

2. In the foor lop, the i value represents the command line argument I need to use, but $$i doesnt work.

./runMe arg1 arg2 outputDir [size1 size2 size3...]

for ((i=4; i<$#; i++ ))
do
    ping -s $$i google.com >> $outputDir/$$iresults.txt
done

I need to know how to build the $outputDir/$$iresults.txt string. Also, the ping -s $$i doesnt work. Its like I need two levels of replacement. I need to replace the $[$i] inner $i with the value in the loop, like 4 for ex, making it $4. Then replace $4 with the 4th command line argument!

Any help would be greatly appreciated.

Thanks!

Answer 1

Indirection uses the ! substitution prefix:

echo "${!i}"

But you should be using a bare for after shifting the earlier ones out:

shift
shift
for f
do
  echo "$f"
done

Answer 2

for ARG in $@; do
  $COMMAND  > ${ARG}.txt
done