pure Bash way to trim

Question

How can I improve this? I added the while read loop to handle preceding newlines.

trim()
{
 # skip leading whitespace
 while read -r string; do [[ -n "${string}" ]] && break; done
 string="${string#"${string%%[![:space:]]*}"}"
 string="${string%"${string##*[![:space:]]}"}"
 printf "${string}"
}

Usage:

$ echo " hello " | trim; echo
hello
$ echo "
 hello multi-line
 " | trim
hello multi-line
$

Answer 1

The comment is misleading - the function does skip leading whitespace-only lines, but it also removes all lines after the first non-whitespace.

I would simplify to read all of the input into string, and then remove the whitespace:

 string=$(cat)
 string="${string#"${string%%[![:space:]]*}"}"

Don't make the result the format-string for printf - any % in the input will be substituted (e.g. trim <<<%d results in 0 as output). Instead, use %s as the format string:

 printf %s "$string"

---

Modified code

trim()
{
 string=$(cat)
 string=${string#${string%%[![:space:]]*}}
 printf %s "${string%${string##*[![:space:]]}}"
}

And some improved tests:

tests=(
 ' hello '
 '
 hello multi-line ''
 '
 '
 two
 lines
 
 '
 '%d'
 ' '
 '
'
)
for s in "${tests[@]}"
do
 printf '%s\n^%s$\n\n' "${s@Q}" "$(trim <<<"$s")"
done