bash parameter-expansion: removing multiple occurences of the same character-CodePudding

It's GNU bash, version 4.4.20(1)-release (x86_64-pc-linux-gnu). I have variable with text in it. For example:

var="this is     a variable   with some text     :)"

now I want to have var2 having the same content but multiple spaces replaced with singe one.
I know we can do it with sed, tr, awk and hundred other ways... but is there a chance to do it with parameter-expansion which bash performs ?
Tried:

var=${var/  / } # replacing 2 spaces with 1, but not much helps
var=${var/[ ]*/ } # this replaces space and whatever follows.... bad idea
var=${var/*( )/}  # found this in man bash, whatever it does it still doesnt help me...

var2=$(echo $var) with hope that echo solves problem - doesnt do the job bcs it doesnt preserve special characters like tab and so on..

I strongly would like to have it solved with something that man bash offers.

CodePudding user response：

*( ) is an extended glob, which needs to be enabled using shopt -s extglob before you can use it to match zero or more spaces.

The correct replacement, though, would be to replace one or more spaces ( ( )) with a single space. You also need to use ${var// ...} to replace every occurrence of multiple spaces, not just the first one.

$ shopt -s extglob
$ echo "$var"
this is     a variable   with some text     :)
$ echo "${var// ( )/ }"
this is a variable with some text :)