I have three arrays:
array1=(8 7 6 5 4 3 2 1)
declare -a array2
declare -a array3
And X that representing which array I should use for some operation So, first of all I am finding it like this:
nameOfArray=array$X[@]
indirectTempArray=("${!nameOfArray}")
echo ${indirectTempArray[@]} // returns 8 7 6 5 4 3 2 1 in case if X == 1
So, the question is, how I can delete value from original array which reference I have?
You can pass a plain string to unset:
array1=(8 7 6 5 4 3 2 1)
X=1
unset "array$X[1]"
declare -p array1
results in the array without the second element (index 1):
declare -a array1=([0]="8" [2]="6" [3]="5" [4]="4" [5]="3" [6]="2" [7]="1")
declare "array$X[42]=1337"declare "array$X+=(42)"If you really want to do it, I would combine the following two ideas:
BashFAQ/06 Evaluating Indirect references
$ realarray=(...); ref=realarray; index=2
$ tmp=${ref}[index]
$ echo "${!tmp}"
delete by key:
$ realarray=( 5 4 3 2 1 "foo bar" ); ref=realarray; key=2
$ tmp=${ref}[@]; tmp=( "${!tmp}" )
$ unset "$ref"'['"$key"']'
$ echo "${realarray[@]}"
5 3 2 1 foo bar
$ echo "${#realarray[@]}"
5
delete by value:
$ realarray=( 5 4 3 2 1 "foo bar" ); ref=realarray; value=2
$ tmp=${ref}[@]; tmp=( "${!tmp}" )
$ eval "$ref=()"
$ for i in "${tmp[@]}"; do [ "$i" != "$value" ] && eval "$ref+=(\"$i\")"; done
$ echo "${realarray[@]}"
5 4 3 1 foo bar
$ echo "${#realarray[@]}"
5
This removes the element from the array and solves quoting issues. You could write this a bit different by using a second temporary and a single eval, but the idea is the same.
