Simple way to get the first non whitespace character from string

The first non whitespace character is a in abc,i get it with:

echo '   abc' | grep -o '^\s\+.' |tr -d  ' '
a

Is there more shorter way to do so?

echo '   abc  sd ggf ' | awk '{print $1}' | cut -c 1
echo '   abc  sd ggf ' | sed 's/^[ \t]*//' | cut -d' ' -f 1 | cut -c 1
echo '   abc  sd ggf ' | sed -r 's/\s+//' | cut -c -1

PS: Corrected for printing the first character.

Retrieving a regex match from the BASH_REMATCH[] array:

str='   abc'
regex='[^[:space:]]'                                          # match a single non-whitespace character
[[ "${str}" =~ ${regex} ]] && char1="${BASH_REMATCH[0]}" 

Result:

$ typeset -p char1
declare -- char1="a"

$ echo "${char1}"
a

NOTE: while this requires a bit more typing it has the benefit of eliminating the performance overhead of creating a subshell for each piped command (eg, | grep, |tr , | awk, | sed, | cut)

Another idea using parameter expansion to strip out whitespace characters plus a substring call:

str='   abc'
newstr="${str//[[:space:]]/}"                 # strip out whitespace
char1="${newstr:0:1}"                         # extract 1st character via substring (start position 0, length of 1)

Result:

$ typeset -p char1
declare -- char1="a"

$ echo "${char1}"
a

NOTE: a little less typing and this also eliminates the performance overhead of creating a subshell for each piped command (eg, | grep, |tr , | awk, | sed, | cut)

Let's shorten your own command first by removing useless spaces and using a here string instead of echo and a pipe. Let's also make it string independent with a bash variable:

s='   abc'
grep -o '^\s\+.'<<<$s|tr -d  ' '

It is now 32 characters long. Can we find shorter? Yes, with tr and cut. If the whitespaces are real spaces (not tabs) here is a 22 characters long script:

tr -d ' '<<<$s|cut -c1

If the whitespaces can also be tabs we need two more characters, that is, 24:

tr -d ' \t'<<<$s|cut -c1

If they can be any horizontal whitespace we need 6 more (30):

tr -d '[:blank:]'<<<$s|cut -c1

And if they can be any horizontal or vertical whitespace (but if the string contains newlines your grep-based solution does not work any more), we still have a 30 characters long script:

tr -d '[:space:]'<<<$s|cut -c1

You can also do it with just parameter expansion forms:

shopt -s extglob # Turn on extended globs to get +()
foo='   abc'
trimmed=${foo##+([[:space:]])} # Remove leading whitespace if any
echo "${trimmed:0:1}" # And display the first character of the remaining text

UPDATE 1 ::: MUCH cleaner

mawk '_ {exit} _ = _$(NF = 1)' RS='[ \t-\r]+' FS=

It assigns the 1st character into _, so as long as the input rows are empty, it wouldn't exit.

The _$( ) isn't anything special - it's merely a straight up string concat of variable _ with field $( … )

>

echo '   abc' | 

gawk '_ {exit} _ = "" < $(NF=1)' RS='[ \t-\r]+' FS= 

a