How to take multiple arguments in bash

Question

I'm trying to make it so my bash shellcode can take multiple files as arguments. How would I do this?

atm this shellcode checks if only one files exsist and it checks every $2 seconds...

How would I make it take more files as arguments?

#!/bin/bash while [ ! -e $1 ] do sleep $2 if [ -e $1 ] then echo "$1 was created!" fi done

How would you make the find command loop until the file is created? — Barmar
– Barmar, Commented Sep 6, 2019 at 2:35

Barmar · Accepted Answer · 2019-09-06 02:38:54Z

Put the sleep time first, then loop over the remaining arguments.

time=$1 shift # Remove time from argument list for file in "$@" do while [ ! -e "$file" ] do sleep "$time" done echo "$file was created!" done

You also don't need to test for the file's existence twice. When the loop ends, you know the file has been created.

If you want to put the time last, you can loop over the argument indexes, using indirect variables.

time=${!#} # last argument for ((i = 1; i < $#; i++)) do file=${!i} while [ ! -e "$file" ] do sleep "$time" done echo "$file was created!" done

Yeah I was thinking of this aswell, but what if I would want the timer at the end?

1 Answer 1