more bash! regex substitution to crop trailing whitespacesy

[Chris F.A. Johnson]

On Tue, 18 Dec 2007, Madison Kelly wrote:

>  I am almost done this darn bash script of mine (insane as it is), but I've
> got one hurdle I can't seem to get over... It *should* be the last hurdle for
> me to finish this things and start the long process of regaining some sanity.
>
>  I read in key=value pairs from a file, which is easy enough. I can cut off
> preceeding white spaces just fine with:
...
>  But I can't get trailing spaces cut off. The '(.*)(\s+)' pattern doesn't 
> match because it's greedy and matches to the end of the string. This:

    This shell function removes leading and trailing spaces (or other
    character if given as the second argument):

_trim ()
{
     _TRIM=$1
     trim_string=${_TRIM%%[!${2:- }]*}
     _TRIM=${_TRIM#"$trim_string"}
     trim_string=${_TRIM##*[!${2:- }]}
     _TRIM=${_TRIM%"$trim_string"}
}

var="    qwerty     "
_trim "$var"
var=$_TRIM
printf ":%s:\n" "$var"

var="......uiop...."
_trim "$var" .
var=$_TRIM
printf ":%s:\n" "$var"


-- 
    Chris F.A. Johnson, webmaster         <http://woodbine-gerrard.com>
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    Author:
    Shell Scripting Recipes: A Problem-Solution Approach (2005, Apress)
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