uppercase first character in a variable with bash

Question

I want to uppercase just the first character in my string with bash.

foo=\"bar\";

//uppercase first character

echo $foo;

should print \"B

Answer 1

Not exactly what asked but quite helpful

declare -u foo #When the variable is assigned a value, all lower-case characters are converted to upper-case.

foo=bar
echo $foo
BAR

And the opposite

declare -l foo #When the variable is assigned a value, all upper-case characters are converted to lower-case.

foo=BAR
echo $foo
bar

Answer 2

Here is the "native" text tools way:

#!/bin/bash

string="abcd"
first=`echo $string|cut -c1|tr [a-z] [A-Z]`
second=`echo $string|cut -c2-`
echo $first$second

Answer 3

just for fun here you are :

foo="bar";    

echo $foo | awk '{$1=toupper(substr($1,0,1))substr($1,2)}1'
# or
echo ${foo^}
# or
echo $foo | head -c 1 | tr [a-z] [A-Z]; echo $foo | tail -c +2
# or
echo ${foo:1} | sed -e 's/^./\B&/'

Answer 4

What if the first character is not a letter (but a tab, a space, and a escaped double quote)? We'd better test it until we find a letter! So:

S='  \"ó foo bar\"'
N=0
until [[ ${S:$N:1} =~ [[:alpha:]] ]]; do N=$[$N+1]; done
#F=`echo ${S:$N:1} | tr [:lower:] [:upper:]`
#F=`echo ${S:$N:1} | sed -E -e 's/./\u&/'` #other option
F=`echo ${S:$N:1}
F=`echo ${F} #pure Bash solution to "upper"
echo "$F"${S:(($N+1))} #without garbage
echo '='${S:0:(($N))}"$F"${S:(($N+1))}'=' #garbage preserved

Foo bar
= \"Foo bar=

Answer 5

foo="$(tr '[:lower:]' '[:upper:]' <<< ${foo:0:1})${foo:1}"

Answer 6

One way with sed:

echo "$(echo "$foo" | sed 's/.*/\u&/')"

Prints:

Bar