This will create arr[0]=hy, arr[1]="hello man":
declare -a arr=(hy "hello man")
I have the 'hy "hello man"' this as whole string stored in variable such that:
echo "$var"
hy "hello man"
How can I split it back to an array? I tried IFS= IFS=" " readarray and all but array only results into arr[0]="hy "hello man"" and not arr[0]=hy, arr[1]="hello man"
Desired array:
arr[0]=hy arr[1]="hello man"
For testing purpose:
arg () { printf "%d args:" "$#"; [ "$#" -eq 0 ] || printf " <%s>" "$@"; echo; }
arg hy "hello man"
2 args: <hy> <hello man>
But
var='hy "hello man"'
arg $t
1 arg: <hy "hello man">
Is there a quoting trick that can help with my problem?
Is there a quoting trick that can help with my problem?
Yes, place the double quotes around the whole array declaration (including the parens):
var='hy "hello man"'
unset arr; declare -a arr="($var)" # or even: declare -a "arr=($var)"
This will restore the desired array (quotes inside $var must be syntactically correct, though):
$ declare -p arr
declare -a arr=([0]="hy" [1]="hello man")
In contrast, putting the qoutes only around the variable would turn it into a single element:
var='hy "hello man"'
unset arr; declare -a arr=("$var")
Resulting in the "wrong" array:
$ declare -p arr
declare -a arr=([0]="hy \"hello man\"")
Tested with GNU bash, version 5.2.37(1)-release (x86_64-pc-linux-gnu)