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
$ foo="bar";
$ foo=`echo ${foo:0:1} | tr '[a-z]' '[A-Z]'`${foo:1}
$ echo $foo
Bar
Linux和OSX的另一种干净的解决方案，它也可以与bash变量一起使用
python -c "print(\"abc\".capitalize())"
返回美国广播公司
foo="$(tr '[:lower:]' '[:upper:]' <<< ${foo:0:1})${foo:1}"
如果第一个字符不是字母(而是制表符、空格和转义双引号)怎么办?我们最好测试它，直到我们找到一个字母!所以:
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=
仅使用awk
foo="uNcapItalizedstrIng"
echo $foo | awk '{print toupper(substr($0,0,1))tolower(substr($0,2))}'
