Bash removing duplicate in a string list of argument $@

user_15586780 0
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In an bash shell script I am receiving a list of argument in $@ which are : a b c d a c e Depending of the argument I need to do something specific with a case structure. But I only want to do it for a b c d e

I should only use bash and not any other language...I can use awk for example

for argument in "$@"
do
    case $argument in
a)
.....

Have tried many things but without success

Any help much appreciated

duplicates bash
2 Answers

Use an associative array to track arguments you've already seen. Note that this requires bash 4.0 or later; the 3.2.x release that Apple ships is too old, as it only supports numerically-indexed arrays (declare -a, but not declare -A).

#!/usr/bin/env bash
case $BASH_VERSION in ''|[0-3].*) echo "ERROR: Bash 4.0+ required" >&2; exit 1;; esac

declare -A seen=( )
declare -a deduped=( )

for arg in "$@"; do                # iterate over our argument list
  [[ ${seen[$arg]} ]] && continue  # already seen this? skip it
  seen[$arg]=1                     # mark as seen going forward...
  deduped+=( "$arg" )              # ...and add to our new/future argv
done

set -- "${deduped[@]}"  # replace "$@" with contents of deduped array
user_141220

You could print, then sort unique, then print quoted the list and re-eval into arguments. With GNU tools:

set -- a b c d a c e
tmp=$(printf "%s\0" "$@" | sort -uz | xargs -0 printf "%q ")
eval set -- "$tmp"
user_90727530
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