getting all elements of a bash array except a given index

Question

Team, I have a use case to save remainder of array into new array while looping on its indices and new array should have all elements except the looped element.

$ a=(1 2 3 4 5)
$ b=()

for index in ${!a[@]}
do
 b=(a - a[$index]) #how would i code his that it works ?
 echo ${b[@]}
done

expected loop values for each iteration.

See [BASH: deleting Nth element in an array without another variable](https://stackoverflow.com/q/71154323/4154375). — pjh, Apr 20 '23 at 00:44

score 2 · Accepted Answer · answered Apr 19 '23 at 22:49

2

Just be simple.

b=("${a[@]}")
unset b[$index]

answered Apr 19 '23 at 22:49

KamilCuk

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See [Bash Pitfalls #57 (unset unquoted array element)](https://mywiki.wooledge.org/BashPitfalls#unset_a.5B0.5D). – pjh Apr 20 '23 at 00:38

score 0 · Answer 2 · answered Apr 19 '23 at 22:59

Use the parameter expansion with the :offset:length syntax:

#!/bin/bash
a=(1 2 3 4 5)
b=()

for index in "${!a[@]}" ; do
    b=("${a[@]:0:index}")

    # This should be "${a[@]:index+1:${#a[@]}-index-1}",
    # but the result is the same.
    b+=("${a[@]:index+1:${#a[@]}}")

    echo "${b[@]}"
done

getting all elements of a bash array except a given index

2 Answers2