用不同字符替换 bash 中的偶数和奇数引号

Question

在 bash 中，使用 sed、awk、ZF83A0AA1F9CA0F7IDD80994445Z7D 等任何常用工具，可以如何转换：

Some line
Some text "quoted" some other text "quoted again"

对此：

Some line
Some text [quoted] some other text [quoted again]

（对不起，忽略颜色堆栈溢出应用于这些代码块。这不是故意的。）

基本上，我希望用两个字符之一替换引号，具体取决于它是行中的偶数还是奇数引号。

我很好，它不能很好地处理“格式不正确”的字符串，例如： This won"t work as intended because "someone" typed double quotes in the word won't instead of a single quote.我不在乎它' 将其更改为This wo[t work as intended because ]someone[ typed double quotes in the word won't instead of a single quote.

（我实际上想要做的是用 ANSI 颜色序列、引用文本、然后是 ANSI 颜色重置（返回白色）序列替换引用文本。但是，我将问题简化为仅在此处使用[和] .)

Answer 1

以下演示代码假定引用部分在它开始的同一行结束。 对于这种情况，应该做一个简单的替换。

use strict;
use warnings;

while( <DATA> ) {
    s/"([^"]+)"/[$1]/g;
    print;
}

__DATA__
Some line
Some text "quoted" some other text "quoted again"

Output

Some line
Some text [quoted] some other text [quoted again]

Answer 2

使用sed ：

$ sed -E 's/"([^"]*)"?/[\1]/g' file
Some text [quoted] some other text [quoted again]
This won[t work as intended because ]someone[ typed double quotes in the word won't instead of a single quote.]

但它会在行尾放一个] 。 这个没有。

$ sed -E 's/"([^"]*)"/[\1]/g ; s/"([^"]*)$/[\1/' file
Some text [quoted] some other text [quoted again]
This won[t work as intended because ]someone[ typed double quotes in the word won't instead of a single quote.