ruby - 如何通过正则表达式查找括号内的文本，但有一些异常(exception)？

Question

我有一个正则表达式 /^\[(text:\s*.+?\s*)\]/mi 目前可以捕获以 text 开头的括号中的文本:。以下是它的工作示例:

[text: here is my text that is
captured within the brackets.]

现在，我想添加一个异常(exception)，以便它允许某些括号，如下例所示:

[text: here is my text that is
captured within the brackets
and also include ![](/some/path)]

基本上，我需要它允许匹配中的 ![](/some/path) 括号。

如有任何帮助，我们将不胜感激。谢谢。

更新:

下面是括号内的文字应该匹配的一些情况:

[text: here is my text that is
captured within the brackets
and also include ![](/some/path)]

[text: here is my text that is
captured within the brackets
and also include ![](/some/path) and some more text]

[text: ![](/some/path)]

![text: cat]

以下是一些不应该匹配的情况:

[text: here is my text that is
captured within the brackets
and also include ![invalid syntax](/some/path)]

[text: here is my text that is
captured within the brackets
and also include ![] (/some/path)]

[text: here is my text that is
captured within the brackets
and also include ! [](/some/path)]

[text: here is my text that is
captured within the brackets
and also include ! [] (/some/path)]

Answer 1

好的，所以你想允许其中之一

• 不是括号或的字符
• 序列![]

在开始和结束括号之间。这给你正则表达式

/^\[(text:[^\[\]]*(?:!\[\][^\[\]]*)*)\]/mi

解释:

^           # Start of line
\[          # Match [
(           # Start of capturing group
 text:      # Match text:
 [^\[\]]*   # Match any number of characters except [ or ]
 (?:        # Optional non-capturing group:
  !\[\]     #  Match ![]
  [^\[\]]*  #  Match any number of characters except [ or ]
 )*         # Repeat as needed (0 times is OK)
)           # End of capturing group
\]          # Match ]

测试一下 live on regex101.com .