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

Answer 1

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

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

Answer 2

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

Answer 3

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

Answer 4

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

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

Answer 5

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 :( )

Answer 6

With cut :

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

Answer 7

Using bash 4:

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