How to obtain the first letter in a Bash variable?

Question

I have a Bash variable, $word, which is sometimes a word or sentence, e.g.:

word="tiger"

Or:

word="This is a sentence."

How can I make a new Bash variable which is equal to only the first letter found in the variable? E.g., the above would be:

echo $firstletter
t

Or:

echo $firstletter
T

Goldsmith

Answer 1

initial="$(echo $word | head -c 1)"

Every time you say "first" in your problem description, head is a likely solution.

Nicky

Please note that there is no -c option in POSIX head: <unix.com/man-page/posix/1/head>. – Dirge

This does return the first letter, but it's an overkill, and the solution depends on external utilities (head). This can be done in pure shell, which is cleaner. – Gorcock

This should not be the accepted answer, for the reasons pointed out by the above comments. The real answers are below - ${word:0:1} or ${word::1}. – Radiator

For POSIX compliance you could do echo "$word" | fold -w1 | head -n 1, printf '%.1s' "$X" or printf '%c' "$X" but these are all subject to single v. multi byte character shenanigans. – Alchemist

Wait, so is head -c n, whoops. echo "fábio" | head -c 2 gives f� , so that's another problem with this I guess. – Alchemist

To update the accepted answer would take a @Villiage. Wanna? The community would thank you ... – Sherris

This is overkill if you want to get the first character(s) of a bash variable. It is a fine solution if you want the first character(s) of anything else regardless of your shell. E.g. in a script run by sh: date +%N | head -c 7. – Newlywed

Answer 2

word="tiger"
firstletter=${word:0:1}

Edna

Exactly what I was looking for. One-liner, without spawning subshells. Take my vote ! – Karlotte

Answer 3

word=something
first=${word::1}

Kassel

This would be wonderful, but does not work for me, neither in bash 4.3.11 nor in zsh 5.0.5 – Echovirus

With GNU bash v4.3.11, first is set to the letter s. You can see it with echo $s. If that doesn't work for you, and you'd like help debugging, please copy and paste the command(s) you typed and their output. – Kassel

Thanks Adam, my bad, I can't figure out why it didn't work for me before, but it works perfectly fine now on several version of bash I tried, going back to GNU bash, v3.1.0. Still doesn't work in zsh, but the question is tagged "bash", so that's irrelevant. – Echovirus

Thanks for following up and confirming! Glad it's working for you. – Kassel

For zsh users: It is working with ${word:0:1} – Marnie

Answer 4

A portable way to do it is to use parameter expansion (which is a POSIX feature):

$ word='tiger'
$ echo "${word%"${word#?}"}"
t

Osmosis

More reasonable to use are the other answers with :: but I can't miss to upvote such an abuse of expansion. Nicely done. – Pet

It's often the case that you need to shell out in a scripting environment that mandates, well POSIX sh, not bash (though that was not part of the question) – Coburg

Answer 5

With cut :

word='tiger'
echo "${word}" | cut -c 1

Oglethorpe

Answer 6

Since you have a sed tag here is a sed answer:

echo "$word" | sed -e "{ s/^\(.\).*/\1/ ; q }"

Play by play for those who enjoy those (I do!):

{

• s: start a substitution routine
  • /: Start specifying what is to be substituted
  • ^\(.\): capture the first character in Group 1
  • .*:, make sure the rest of the line will be in the substitution
  • /: start specifying the replacement
  • \1: insert Group 1
  • /: The rest is discarded;
• q: Quit sed so it won't repeat this block for other lines if there are any.

}

Well that was fun! :) You can also use grep and etc but if you're in bash the ${x:0:1} magick is still the better solution imo. (I spent like an hour trying to use POSIX variable expansion to do that but couldn't :( )

Alchemist

Answer 7

Using bash 4:

x="test"
read -N 1 var <<< "${x}"
echo "${var}"

Dysfunction