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
$ foo="bar";
$ foo=`echo ${foo:0:1} | tr '[a-z]' '[A-Z]'`${foo:1}
$ echo $foo
Bar
其他回答
它也可以在纯bash中使用bash-3.2完成:
# First, get the first character.
fl=${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}"
这个对我很有用:
搜索当前目录下的所有*php文件，并将每个文件名的第一个字符替换为大写字母:
例如:test.php => test.php
for f in *php ; do mv "$f" "$(\sed 's/.*/\u&/' <<< "$f")" ; done
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 ${foo:1} | sed -e 's/^./\B&/'
Posix兼容，子进程更少:
v="foo[Bar]"
printf "%s" "${v%"${v#?}"}" | tr '[:lower:]' '[:upper:]' && printf "%s" "${v#?}"
==> Foo[Bar]
使用bash(版本4+)的一种方法:
foo=bar
echo "${foo^}"
打印:
Bar
