如何使用NSRegularExpression删除字符串中的括号词? [英] How do you remove parentheses words within a string using NSRegularExpression?

问题描述

我对regex不太熟悉，因此我一直想让它与Apple的NSRegularExpression一起使用

我正在尝试删除括号或方括号中的单词...

例如:

NSString * str = @如何在字符串中使用<> (删除括号中的单词)

结果字符串应为:@如何在字符串中使用"

谢谢！！

解决方案

搜索

\([^()]*\)

然后什么也不要替换.

作为详细的正则表达式:

\(      # match an opening parenthesis
[^()]*  # match any number of characters except parentheses
\)      # match a closing parenthesis

如果括号正确平衡且未嵌套，这将很好地工作.如果括号可以嵌套(like this (for example))，则您需要重新运行替换操作，直到没有其他匹配项为止，因为每次运行仅会匹配最里面的括号.*

要卸下括号，请对\[[^[\]]*\]进行相同的操作，以将括号\{[^{}]*\}括起来.

使用条件表达式，您可以一次完成所有三个操作，但是正则表达式看起来很丑，不是吗?

(?:(\()|(\[)|(\{))[^(){}[\]]*(?(1)\))(?(2)\])(?(3)\})

但是，我不确定NSRegularExpression是否可以处理条件语句.可能不是.这个怪物的解释:

(?:           # start of non-capturing group (needed for alternation)
 (\()         # Either match an opening paren and capture in backref #1
 |            # or
 (\[)         # match an opening bracket into backref #2
 |            # or
 (\{)         # match an opening brace into backref #3
)             # end of non-capturing group
[^(){}[\]]*   # match any number of non-paren/bracket/brace characters
(?(1)\))      # if capturing group #1 matched before, then match a closing parenthesis
(?(2)\])      # if #2 matched, match a closing bracket
(?(3)\})      # if #3 matched, match a closing brace.

* _{您不能将任意嵌套的括号(因为这些构造不再是正则表达式)与正则表达式进行匹配，因此，这不仅不是此regex的限制，而是一般的正则表达式的限制.}

I am not too familiar with regex and so I been having some getting this to work with Apple's NSRegularExpression

I am trying to remove words in parentheses or brackets...

For example:

NSString *str = @"How do you (remove parentheses words) within a string using"

resulting string should be: @"How do you within a string using"

Thanks you!!!

解决方案

Search for

\([^()]*\)

and replace with nothing.

As a verbose regex:

\(      # match an opening parenthesis
[^()]*  # match any number of characters except parentheses
\)      # match a closing parenthesis

This will work fine if parentheses are correctly balanced and unnested. If parentheses can be nested (like this (for example)), then you need to re-run the replace until there are no further matches, since only the innermost parentheses will be matched in each run.*

To remove brackets, do the same with \[[^[\]]*\], for braces \{[^{}]*\}.

With conditional expressions you could do all three at once, but the regex looks ugly, doesn't it?

(?:(\()|(\[)|(\{))[^(){}[\]]*(?(1)\))(?(2)\])(?(3)\})

However, I'm not sure if NSRegularExpression can handle conditionals. Probably not. Explanation of this monster:

(?:           # start of non-capturing group (needed for alternation)
 (\()         # Either match an opening paren and capture in backref #1
 |            # or
 (\[)         # match an opening bracket into backref #2
 |            # or
 (\{)         # match an opening brace into backref #3
)             # end of non-capturing group
[^(){}[\]]*   # match any number of non-paren/bracket/brace characters
(?(1)\))      # if capturing group #1 matched before, then match a closing parenthesis
(?(2)\])      # if #2 matched, match a closing bracket
(?(3)\})      # if #3 matched, match a closing brace.

*_{You can't match arbitrarily nested parentheses (since these constructs are no longer regular) with regular expressions, so that's not a limitation of this regex in particular but of regexes in general.}

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