用 sed 删除大括号对之间的所有内容 [英] Remove everything between pairs of braces with sed

问题描述

我有一个看起来像这样的字符串:

I've got a string that looks like this:

[%{%B%F{blue}%}master %{%F{red}%}*%{%f%k%b%}%{%f%k%b%K{black}%B%F{green}%}]

我想删除与 %{...} 匹配的子字符串，这些子字符串可能包含也可能不包含相同顺序的其他子字符串.

I want to remove the substrings matching %{...}, which may or may not contain further substrings of the same order.

我应该得到:[master *] 作为最终输出.我目前的进展:

I should get: [master *] as the final output. My progress so far:

gsed -E 's/%\{[^\}]*\}//g'

给出:

echo '[%{%B%F{blue}%}master %{%F{red}%}*%{%f%k%b%}%{%f%k%b%K{black}%B%F{green}%}]' | gsed -E 's/%\{[^\}]*\}//g'
[%}master %}*%B%F{green}%}]

因此，这适用于 %{...} 部分，这些部分不包含 %{...}.对于像 %{%B%F{blue}%} 这样的字符串，它会失败(它返回 %}).

So, this works fine for %{...} sections which do not contain %{...}. It fails for strings like %{%B%F{blue}%} (it returns %}).

我想要做的是解析字符串，直到找到匹配 }，然后删除到该点的所有内容，而不是删除 % 之间的所有内容{ 和我遇到的第一个 }.我不知道该怎么做.

What I want to do is parse the string until I find the matching }, then remove everything up to that point, rather than removing everything between %{ and the first } I encounter. I'm not sure how to do this.

我完全意识到可能有多种方法可以做到这一点；如果可能的话，我更喜欢关于问题中指定方式的答案，但任何想法都非常受欢迎.

I'm fully aware that there are probably multiple ways to do this; I'd prefer an answer regarding the way specified in the question if it is possible, but any ideas are more than welcome.

推荐答案

这可能对你有用:

echo '[%{%B%F{blue}%}master %{%F{red}%}*%{%f%k%b%}%{%f%k%b%K{black}%B%F{green}%}]' |
sed 's/%{/{/g;:a;s/{[^{}]*}//g;ta'
[master *]

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