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Sed（或其他）脚本用于替换捕获组内的字符

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Sed（或其他）脚本用于替换捕获组内的字符

我正在尝试将 Pandoc 标记转换为 Confluence wiki 标记，我正在使用markdown2confluence来完成大部分工作。除了我谈论的 CSS 和 FreeMarker在代码中使用{&而 Confluence 使用&来标记代码块的开始/结束之外，这种方法效果很好。所以我需要匹配一个包含在 中的模式。}{{}}{{...}}

如果我了解更多 Ruby，我可能会在那里修复它，但我是一个老派 Unix 人，所以我想到了 awk 或 sed。

因此我得到了：

   sed 's/{{\([^}}]*\)}}/{{"\1"}}/g' tmp.wkd

需要：

First we need a way to select a state (or group of states) CSS uses what
is called a selector to choose which elements to apply to, we have been
using one up until now without noticing, it is the {{*}} at the beginning
of our CSS. This is a special selector that means select everything. So
the rule that follows it (the bit between {{{}} and {{}}} apply to every
polygon on the map. But CSS allows us to insert a filter instead by
using {{[...]}} instead of {{*}}.

并产生：

First we need a way to select a state (or group of states) CSS uses what
is called a selector to choose which elements to apply to, we have been
using one up until now without noticing, it is the {{"*"}} at the beginning
of our CSS. This is a special selector that means select everything. So
the rule that follows it (the bit between {{"{"}} and {{""}}} apply to every
polygon on the map. But CSS allows us to insert a filter instead by
using {{"[...]"}} instead of {{"*"}}.

但我需要的是：

First we need a way to select a state (or group of states) CSS uses what
is called a selector to choose which elements to apply to, we have been
using one up until now without noticing, it is the {{*}} at the beginning
of our CSS. This is a special selector that means select everything. So
the rule that follows it (the bit between {{\{}} and {{\}}} apply to every
polygon on the map. But CSS allows us to insert a filter instead by
using {{[...]}} instead of {{*}}.

还需要处理{{${type.name}}}应该成为的{{$\{type.name\}}}。

有两个问题

  1. 我需要{用 来替换\{，而不是使用引号，所以我需要以\1某种方式进行修改。
  2. 无论我如何尝试结束模式匹配，看起来令人讨厌{{}}}（应该是出现）并没有正确出现。{{\}}}

答案1

以下 sed 命令似乎有效：

   sed 's/{{\([^*[a-z][^}]*\)}}/{{\\\1}}/g;s/{{\\${\([^}]*\)}}}/{{$\\{\1\\}}}/g'

解释：

  1. {{\([^*[a-z][^}]*\)}}查找{{stuff}}，除非stuff以*或[或小写字母开头。
  2. 将其替换为{{\stuff}}。
  3. 然后{{\\${\([^}]*\)}}}发现{{\${junk}}}。
  4. 并将其替换为{{$\{junk\}}}。

编辑：在 OP 澄清之后，另一种解决方案可能是：

   sed 's/\({{[^}]*\){\([^}]*}}\)/\1\\{\2/g;s/\({{[^}]*\)}}}/\1\\}}}/g'

众所周知，sed 不能进行递归解析，但这对于大多数简单情况应该有效。

解释：

  1. \({{[^}]*\){\([^}]*}}\)发现{{foo{bar}}，其中foo和bar不包含}。
  2. 并将其替换为{{foo\{bar}}。（注释{{xxx{yyy}}}处理正确。）
  3. 然后\({{[^}]*\)}}}发现{{baz}}}，其中baz不包含}。
  4. 并将其替换为{{baz\}}}。

foo、bar和baz可以为空，因此例如根据需要{{}}}转换为{{\}}}。

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