Question

我想捕捉匹配的花括号。

例如：

<script src="https://ajax.googleapis.com/ajax/libs/jquery/1.9.1/jquery.min.js"></script> <div id="prf"> <label id='lblreason'></label> <input type='text' id='reason' /> <label id='lbldate'></label> <input type='date' id='date' placeholder='mm/dd/yyyy'> </div> <select id="status"> <option>Outstanding</option> <option disabled=''>Served</option> <option>Cancelled</option> <option>Hold</option> </select>

Some example text with \added[author]{text with curly braces{some text}..}

Some example text with \added[author]{text without curly braces}

Some example text with \added[author]{text with {}and {} and {}curly braces{some text}..}

预期输出：

Some example text with \added[author]{text with {}and {} and {}curly braces{some text}..} and extented text with curly braces {}

Some example text with text with curly braces{some text}..

Some example text with text without curly braces

Some example text with text with {}and {} and {}curly braces{some text}..

即。我想捕捉Some example text with text with {}and {} and {}curly braces{some text}.. and extented text with curly braces {}和\added[]{之间的文本（它的相对结束花括号）。我的正则表达式的问题是，我不知道如何捕捉相关的花括号。

我试过了，

我知道如果文本中出现"/\\\\added\\[.*?\\]{(.[^{]*?)}/s"则忽略它。但我不知道如何创建一个正则表达式来单独匹配花括号。

Answer 1

要匹配成对的大括号，您需要使用recursive subpattern。

示例：

$regex = <<<'REGEX' / \\added\[.*?\] # Initial \added[author] ( # Group to be recursed on. { # Opening brace. ( # Group for use in replacement. ((?>[^{}]+)|(?1))* # Any number of substrings which can be either: # - a sequence of non-braces, or # - a recursive match on the first capturing group. ) } # Closing brace. ) /xs REGEX; $strings = [ 'Some example text with \added[author]{text with curly braces{some text}..}', 'Some example text with \added[author]{text without curly braces}', 'Some example text with \added[author]{text with {}and {} and {}curly braces{some text}..}', 'Some example text with \added[author]{text with {}and {} and {}curly braces{some text}..} and extented text with curly braces {}' ]; foreach ($strings as $string) { echo preg_replace($regex, '$2', $string), "\n"; }

<强>输出：

Some example text with text with curly braces{some text}.. Some example text with text without curly braces Some example text with text with {}and {} and {}curly braces{some text}.. Some example text with text with {}and {} and {}curly braces{some text}.. and extented text with curly braces {}

Answer 2

在这里，应该工作

/\\added\[.*\]\{(.*(?:.*\{.*\}.*)*)\}/gU

<强>解释

/\\added\是Latex标签，

\[.*\]是Latex标签的选项，

\{开括号，

(.*(?:.*\{.*\}.*)*)是捕获的文本，我们在此处也阻止了目标代码中的递归{...}或多个{...}，

\}关闭括号。

<强>策略

我不认为对括号是一种递归形式

{ { {...} } }
c b a   a b c

我们有a，b和c，

但我认为他们是这样的！

{ { {...} } }   
a b c   a b c

请参阅：DEMO

我的演示中的最后两个示例也证明它可以正常工作。

重要：修饰符U假设在这里用于非贪婪量词的目的，否则我的正则表达式将无法正常工作。

Answer 3

使用以下正则表达式：

\\\\added\\[[^\\]]\*][^\\{]\*{((?:(?:[^\\{\\}]\*\\{[^\\}\\{]\*\\})\*||[^\\}]\*)\*)}

Answer 4

使用此 regex

/\\added[^]]*]{([^}]*}[^}]*)}/s

<强> Demo here

正则表达式：捕获成对的花括号

4 个答案: