How to obtain the first letter in a Bash variable?
BashText ProcessingVariable ExpansionBash Problem Overview
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 $firstletter
T
Bash Solutions
Solution 1 - Bash
word="tiger"
firstletter=${word:0:1}
Solution 2 - Bash
word=something
first=${word::1}
Solution 3 - Bash
initial="$(echo $word | head -c 1)"
Every time you say "first" in your problem description, head is a likely solution.
Solution 4 - Bash
A portable way to do it is to use parameter expansion (which is a POSIX feature):
$ word='tiger'
$ echo "${word%"${word#?}"}"
t
Solution 5 - Bash
With cut :
word='tiger'
echo "${word}" | cut -c 1
Solution 6 - Bash
Since you have a sed tag here is a sed answer:
echo "$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: Quitsedso 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 ${x:0:1} magick is still the better solution imo. (I spent like an hour trying to use POSIX variable expansion to do that but couldn't :( )
Solution 7 - Bash
Using bash 4:
x="test"
read -N 1 var <<< "${x}"
echo "${var}"