How to trim whitespace from a Bash variable?

Question

I have a shell script with this code:

var=`hg st -R \"$path\"`
if [ -n \"$var\" ]; then
    echo $var
fi

But the conditional code always ex

Answer 1

There are a lot of answers, but I still believe my just-written script is worth being mentioned because:

• it was successfully tested in the shells bash/dash/busybox shell
• it is extremely small
• it doesn't depend on external commands and doesn't need to fork (->fast and low resource usage)
• it works as expected:
  • it strips all spaces and tabs from beginning and end, but not more
  • important: it doesn't remove anything from the middle of the string (many other answers do), even newlines will remain
  • special: the "$*" joins multiple arguments using one space. if you want to trim & output only the first argument, use "$1" instead
  • if doesn't have any problems with matching file name patterns etc

The script:

trim() {
  local s2 s="$*"
  until s2="${s#[[:space:]]}"; [ "$s2" = "$s" ]; do s="$s2"; done
  until s2="${s%[[:space:]]}"; [ "$s2" = "$s" ]; do s="$s2"; done
  echo "$s"
}

Usage:

mystring="   here     is
    something    "
mystring=$(trim "$mystring")
echo ">$mystring<"

Output:

>here     is
    something<