Question

我有这样的事情：

$arr[] = 'Seto Hakashima'
$arr[] = 'Anna (segment "Yvan Attal") (as Robin Wright Penn)'
$arr[] = 'Sara (segment "Yvan Attal")'

我需要删除第二对括号（仅当有第二对时），并得到这个：

$arr[] = 'Seto Hakashima'
$arr[] = 'Anna (segment "Yvan Attal")'
$arr[] = 'Sara (segment "Yvan Attal")'

谢谢！

Answer 1

尝试

preg_replace('/^([^(]+(?:\([^)]+\))?).*/','$1', $item);

几点解释

^          - start of the string
[^(]+      - match characters before first bracket
\([^)]+\)  - match first bracket
(?: ... )? - optional
.*         - eat the rest
$1         - replace with match string

或者只删除最后一部分

preg_replace('/(?<=\))\s*\(.*$/','', $item);

(?<=\))    - if there is ) before pattern
(\s*\(.*$  - remove everything after `(` and also zero or more whitespaces before last bracket.

Answer 2

这有效：

<?php
$arr[] = 'Seto Hakashima';
$arr[] = 'Anna (segment "Yvan Attal") (as Robin Wright Penn)';
$arr[] = 'Sara (segment "Yvan Attal")';
$arr[] = 'Anna (segment "Yvan Attal") (as Robin Wright Penn) BONUS text after second group';

foreach ($arr as $item) {
    print preg_replace('/(\([^\)]*\)[^\(]+)\([^\)]*\)\s*/','$1',$item) . "\n";
}

输出：

Seto Hakashima

Anna（段“Yvan Attal”）

Sara（段“Yvan Attal”）

Anna（段“Yvan Attal”）第二组之后的奖励文本

正如你在上一个例子中所注意到的，这个正则表达式足够具体，它只消除了第二组括号，并保持了其余的字符串。

正则表达式删除第二个括号（仅在可用时）

2 个答案: