regex - 如果第一个字母是大写，则使用 sed 替换字符串

Question

我尝试写这个问题的 sed 答案使用 sed/awk 编辑文件：

sed -e 's/^[A-Z]/$:$&/' file.txt

但结果是：

wednesday
$:$Weekday

$:$thursday
$:$Weekday

$:$friday
$:$Weekday

$:$saturday
$:$MaybeNot

$:$sunday
$:$MaybeNot

$:$monday
$:$Weekday

$:$tuesday
$:$Weekday

如果第一个字符是小写，为什么要替换？

Answer 1

根据此错误报告，这是由语言环境中意外字符排序引起的“功能”，在此处和此处进一步解释。

$ locale
LANG=en_GB.UTF-8
LC_CTYPE="en_GB.UTF-8"
LC_NUMERIC="en_GB.UTF-8"
LC_TIME="en_GB.UTF-8"
LC_COLLATE="en_GB.UTF-8"
LC_MONETARY="en_GB.UTF-8"
LC_MESSAGES="en_GB.UTF-8"
LC_ALL=

$ echo "ABCDEFGHIJKLMNOPQRSTUVWXYZabcdefghijklmnopqrstuvwxyz" | sed -e 's/[A-Z]/./g'
..........................a.........................

$ echo "ABCDEFGHIJKLMNOPQRSTUVWXYZabcdefghijklmnopqrstuvwxyz" | sed -e 's/[a-z]/./g'
.........................Z..........................

$ echo "ABCDEFGHIJKLMNOPQRSTUVWXYZabcdefghijklmnopqrstuvwxyz" | LC_ALL=C sed -e 's/[A-Z]/./g'
..........................abcdefghijklmnopqrstuvwxyz

$ echo "ABCDEFGHIJKLMNOPQRSTUVWXYZabcdefghijklmnopqrstuvwxyz" | LC_ALL=C sed -e 's/[a-z]/./g'
ABCDEFGHIJKLMNOPQRSTUVWXYZ..........................

$ echo "ABCDEFGHIJKLMNOPQRSTUVWXYZabcdefghijklmnopqrstuvwxyz" | sed -e 's/[[:upper:]]/./g'
..........................abcdefghijklmnopqrstuvwxyz

$ echo "ABCDEFGHIJKLMNOPQRSTUVWXYZabcdefghijklmnopqrstuvwxyz" | sed -e 's/[[:lower:]]/./g'
ABCDEFGHIJKLMNOPQRSTUVWXYZ..........................

$ sed --version
GNU sed version 4.2.1