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 Error, Please provide at least one type, ex: sizeofTypes {filetype1} {filetype2}"
  exit 0
fi

#set first filetype
types="-name *."$1

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

echo "TYPES="$types

find . -name '*.'$1 | xargs du -ch *.$1 | grep total

The problem I'm having is right here:

 #loop through additional filetypes and append
    num=1
    while [ $num -lt $# ]
    do
      (( num++ ))
      types=$types' -o -name *.'>>$$num<<
    done

I simply want to iterate over all the arguments not including the first one, should be easy enough, but I'm having a difficult time figuring out how to make this work

Zac Thompson · Accepted Answer · 2010-04-05 23:05:24Z

17

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

answered Apr 5, 2010 at 23:05

Zac Thompson

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Comments

amertune · Accepted Answer · 2010-04-05 23:26:42Z

6

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

answered Apr 5, 2010 at 23:26

amertune

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3 Comments

Corbin Tarrant Over a year ago

this works, but I don't understand why the ! is required, it seems like ${num} should work. How would you read that statment? looks like "not num" but that doesn't make sense

amertune Over a year ago

${num} is the same as $num. ${!num} is the notation for an indirect variable. It looks like \$$num would also work, although, according to what I've read, "The ${!variable} notation is greatly superior to the old 'eval var1=\$$var2'" (faqs.org/docs/abs/HTML/bash2.html)

Sanford Freedman Over a year ago

Should the -le (less than or equal to) be -lt (less than)? When using -le, the loop executes one time more than desired, resulting in use of an unbound variable.

ghostdog74 · Accepted Answer · 2010-04-05 23:44:01Z

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.

answered Apr 5, 2010 at 23:44

ghostdog74

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Comments

Sanford Freedman · Accepted Answer · 2023-03-11 13:43:03Z

0

Here is a non-destructive version that allows for accessing the 1st argument when errexit is enabled:

#!/usr/bin/env bash
set -o nounset -o pipefail -o errexit

Test () {
    local INDEX=0
    while [ $INDEX -lt $# ]
    do
        INDEX=$(( INDEX+1 ))
        echo "->${!INDEX}<-"
    done
}

Test a b c 1 2 3

Output:

->a<-
->b<-
->c<-
->1<-
->2<-
->3<-

This approach does not use shift, allowing argument reuse. Other answers use (( INDEX++ )) to advance the index, but when INDEX is initially set to 0, (( INDEX++ )) increments INDEX, but evaluates to 0, which Bash considers an error.

edited Mar 11, 2023 at 13:43

answered Mar 10, 2023 at 15:12

Sanford Freedman

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