Question

bash中是否有办法获取一定长度的子字符串（在终端中占用＃字符），同时保留颜色代码？

我想我可以用一些代码解释我的意思更好（假装粗体意味着绿色）：

$ # One string is green and another is normal.
$ green_string="\e[1;32mFoo bar baz buzz.\e[0m"
$ normal_string=`echo -e $green_string | sed -r "s/\x1B\[([0-9]{1,2}(;[0-9]{1,2})?)?[mGK]//g"`
$ 
$ # Echo both of them out to show the difference
$ echo -e $green_string
Foo bar baz buzz.
$ echo -e $normal_string
Foo bar baz buzz.
$ 
$ # Get a substring of both.
$ green_sub=${green_string:0:11}
$ normal_sub=${normal_string:0:11}
$ 
$ # Since the special characters are a part of the string, the colored
$ # substring (rightfully) has fewer printable characters in it.
$ echo -e "$green_sub\e[0m"
Foo
$ echo -e $normal_sub
Foo bar baz
$ 
$ # Is there any built in way of doing something like this:
$ green_sub=`some_method $green_string 11`
$ normal_sub=`some_method $normal_string 11`
$ echo -e "$green_sub\e[0m"; echo -e $normal_sub
Foo bar baz
Foo bar baz

对于像我这样的人来说，这是一个复制/粘贴版本：

green_string="\e[1;32mFoo bar baz buzz.\e[0m"
normal_string=`echo -e $green_string | sed -r "s/\x1B\[([0-9]{1,2}(;[0-9]{1,2})?)?[mGK]//g"`
echo -e $green_string
echo -e $normal_string
green_sub=${green_string:0:11}
normal_sub=${normal_string:0:11}
echo -e "$green_sub\e[0m"
echo -e $normal_sub
# green_sub=`some_method $green_string 11`
# normal_sub=`some_method $normal_string 11`
# echo -e "$green_sub\e[0m"; echo -e $normal_sub

我创建了一个函数，为了复制/粘贴演示目的，获取ls的输出并使其恰好填充终端的一行（或更短）：

function lsline {
    color_out=$( ls --color=always | tr "\n" " " )
    max_len=`tput cols`
    cur_len=0
    is_esc=0
    # This is the build of the final string that will be printed out.
    build=""
    # This is the build of any escape sequence so I know not to 
    # include it in the substring count
    esc_build=""
    for (( i=0; i<${#color_out}; i++ )); do
        char="${color_out:$i:1}"
        # Is there an easier way to check if a char is an escape character?
        code=$( printf %x "'$char" )

        # Currently in an escape sequence, so don't count towards max length.
        if [ "$is_esc" -eq "1" ]; then
            esc_build="$esc_build$char"
            if [ "$char" = "m" ]; then
                is_esc=0
                build="$build$esc_build"
                esc_build=""
            fi
        elif [ "$code" = "1b" ]; then
            # 27 is escape character.
            is_esc=1
            esc_build="$char"
        else
            # Not an escape sequence, so build our normal string and update
            # the current substring length.
            build="$build$char"
            ((cur_len++))
            if [ "$cur_len" -eq "$max_len" ]; then
                build="$build$( tput sgr0 )"
                break
            fi
        fi
    done

    echo "$build"
}

此代码有效，但 soooo slow 。

请告诉我有更快/更简单的方法！

Answer 1

使用正则表达式跳过ANSI序列，仅捕获“普通”文本。下面是一个正则表达式的示例，它捕获（可见）字符串的前5个字符，应用于常规字符串和绿色字符串。

$ regex='\e\[[^m]*m(.{5}).*\e\[0m'
$ for str in '\e[1;32mFoo bar baz\e[0m' 'Foo bar baz'; do
> [[ $str =~ $regex ]] && substr=${BASH_REMATCH[1]}
> echo "$substr"
> done
Foo b
Foo b

Bash - 使用颜色代码获取子字符串

1 个答案: