regex - 为匹配中的每一行添加前缀的正则表达式

Question

我想重新格式化我的 Swift 文档

/**
  Lorem Ipsum is simply dummy text of the printing and typesetting industry. 
  Lorem Ipsum has been the industry's standard dummy text ever since the 
  1500s, when an unknown printer took a galley of type and scrambled it to 
  make a type specimen book. It has survived not only five centuries, but 
  also the leap into electronic typesetting, remaining essentially 
  unchanged. It was popularised in the 1960s with the release of Letraset 
  sheets containing Lorem Ipsum passages, and more recently with desktop 
  publishing software like Aldus PageMaker including versions of Lorem Ipsum.
  */

到

/// Lorem Ipsum is simply dummy text of the printing and typesetting industry. 
/// Lorem Ipsum has been the industry's standard dummy text ever since the 
/// 1500s, when an unknown printer took a galley of type and scrambled it to 
/// make a type specimen book. It has survived not only five centuries, but 
/// also the leap into electronic typesetting, remaining essentially 
/// unchanged. It was popularised in the 1960s with the release of Letraset 
/// sheets containing Lorem Ipsum passages, and more recently with desktop 
/// publishing software like Aldus PageMaker including versions of Lorem Ipsum.

使用 Xcode 的查找/替换正则表达式引擎的正则表达式查找和替换 edit: 将是首选，因为只有上帝知道我的 200 多个不同类的库中有多少文档注释。查找这些 block 的表达式很简单，但我不知道如何制作替换表达式以便能够为每一行添加前缀。

当前搜索表达式: (?s:/\*\*(.*?)\*/) - 匹配 /* 之间的所有文本* */

当前替换表达式:///$1

显然，上面的表达式并不能完全达到我的要求。我提前感谢任何帮助!谢谢。

Answer 1

这是我能想到的最好的解决方案，但它依赖于一些专门的 PCRE anchor (最重要的是 \G ，但我也使用了 \A 和 \K )所以它可能不适合你的口味的正则表达式。这也是一个两步解决方案，但我认为不可能将其简化为一个步骤——很想看到有人在这里证明我错了!

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首先，您要匹配 /** 之间以空格开头的每一行和 */并将空格替换为 /// .

查找:

~(?<=/[*]{2}|(?<!\A)\G)\n\K^\s*+(?![*]/)(.*)$~gm

替换:

/// $1

Demo.

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然后我们要获取替换的结果并删除表示旧注释的剩余行，/**和 */ .

查找:

~^.*(?:/[*]{2}|[*]/).*$\n?~gm

替换:

[null]

Demo.

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上面的大部分内容都是直截了当的，或者至少应该假设您对正则表达式有基本的了解……看起来您确实了解。最明显的是第一个。让我们把它分解(耶!)...

(?<=       # start a zero-length lookahead
  /[*]{2}  # look for the start of a comment
 |         # or...
  (?<!\A)  # negate this part if we're at the beginning of the string
  \G       # start at the end of the last match (or beginning of string)
)          # end that lookahead and move on to each line
\n\K       # find the newline and then reset the match for clarity
^\s*+      # match whitespace at the beginning of the line
(?![*]/)   # negate this match if we're at the end of the comment
(.*)$      # capture everything up until the end of the line

上面唯一可能需要额外解释的是(?<!\A)\G)。 “hack”我用。 \G让我们在最后一场比赛结束时开始一场比赛，这在我们这里遇到的重复问题中是非常必要的(对于像这样的包罗万象的解决方案)。然而，\G也匹配我们不想要的字符串的开头(我们在前瞻的前半部分处理它，我们匹配注释的开头)所以我们否定字符串开头与 (?<!\A) 的匹配。 .轰!

\K不是必需的，但是当我们变得如此复杂时，可以使表达更清晰。没有它，\n是比赛的一部分，我们需要手动将其替换为 \n\\\ $1相反。