我想用bash将字符串中的第一个字符大写。
foo="bar";
//uppercase first character
echo $foo;
应打印“Bar”;
当前回答
它也可以在纯bash中使用bash-3.2完成:
# First, get the first character.
fl=${foo:0:1}
# Safety check: it must be a letter :).
if [[ ${fl} == [a-z] ]]; then
# Now, obtain its octal value using printf (builtin).
ord=$(printf '%o' "'${fl}")
# Fun fact: [a-z] maps onto 0141..0172. [A-Z] is 0101..0132.
# We can use decimal '- 40' to get the expected result!
ord=$(( ord - 40 ))
# Finally, map the new value back to a character.
fl=$(printf '%b' '\'${ord})
fi
echo "${fl}${foo:1}"
其他回答
仅使用awk
foo="uNcapItalizedstrIng"
echo $foo | awk '{print toupper(substr($0,0,1))tolower(substr($0,2))}'
这里只是为了好玩:
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&/'
虽然不是我要求的，但很有帮助
declare -u foo #When the variable is assigned a value, all lower-case characters are converted to upper-case.
foo=bar
echo $foo
BAR
反之亦然
declare -l foo #When the variable is assigned a value, all upper-case characters are converted to lower-case.
foo=BAR
echo $foo
bar
使用sed的一种方法:
echo "$(echo "$foo" | sed 's/.*/\u&/')"
打印:
Bar
foo="$(tr '[:lower:]' '[:upper:]' <<< ${foo:0:1})${foo:1}"
