Looping through arguments in Shell Scripting skipping the first argument

Question

#!/bin/sh
BACKUPDIR=$1
for argnum in {2..$#};do
    echo ${"$argnum"}
done

I have tried this but it gives me this error: ./backup.sh: 10: ./backup.sh: Bad substitution

BTW, all-caps names are used for variables with meaning to the OS or shell; variables you define yourself should have at least one lower-case character in their names. See paragraph 4 of the relevant POSIX spec at pubs.opengroup.org/onlinepubs/9699919799/basedefs/… — Charles Duffy
– Charles Duffy, Commented Feb 24, 2018 at 17:19
(Also, you can't use variables in a brace expansion such as {2..$#} -- see BashPitfalls #33). — Charles Duffy
– Charles Duffy, Commented Feb 24, 2018 at 17:20
...actually, brace expansion for numeric ranges isn't guaranteed to work with /bin/sh at all, even with constants. — Charles Duffy
– Charles Duffy, Commented Feb 24, 2018 at 17:25

Charles Duffy · Accepted Answer · 2018-02-24 17:23:19Z

2

Use the shift command to remove $1 from the argument list after you're done reading it (thus renumbering your old $2 to $1, your old $3 to $2, etc):

#!/bin/sh
backupdir=$1; shift
for arg; do
  echo "$arg"
done

To provide a literal (but not-particularly-good-practice) equivalent to the code in the question, indirect expansion (absent such security-impacting practices as eval) looks like the following:

#!/bin/bash
#      ^^^^-- This IS NOT GUARANTEED TO WORK in /bin/sh

# not idiomatic, not portable to baseline POSIX shells; this is best avoided
backupdir=$1
for ((argnum=2; argnum<=$#; ++argnum)); do
  echo "${!argnum}"
done

edited Feb 24, 2018 at 17:23

answered Feb 24, 2018 at 17:18

Charles Duffy

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