How do I use a variable argument number in a bash script?

Question

#!/bin/bash
# Script to output the total size of requested filetype recursively

# Error out if no file types were provided
if [ $# -lt 1 ]
then 
  echo \"Syntax Err         


        

Answer 1

if you don't want to include the first one, the way to do that is to use shift. Or you can try this. imagine variable s is your arguments passed in.

$ s="one two three"
$ echo ${s#* }
two three

Of course, this assume you won't be passing in strings that is one word by itself.

Answer 2

from the bash man page:

  shift [n]
          The  positional  parameters  from n+1 ... are renamed to $1 ....
          Parameters represented by the numbers  $#  down  to  $#-n+1  are
          unset.   n  must  be a non-negative number less than or equal to
          $#.  If n is 0, no parameters are changed.  If n is  not  given,
          it  is assumed to be 1.  If n is greater than $#, the positional
          parameters are not changed.  The return status is  greater  than
          zero if n is greater than $# or less than zero; otherwise 0.

So your loop is going to look something like this:

#loop through additional filetypes and append
while [ $# -gt 0 ]
do
  types=$types' -o -name *.'$1
  shift
done

Answer 3

If all you're trying to do is loop over the arguments, try something like this:

for type in "$@"; do
    types="$types -o -name *.$type"
done

To get your code working though, try this:

#loop through additional filetypes and append
num=1
while [ $num -le $# ]
do
    (( num++ ))
    types=$types' -o -name *.'${!num}
done