删除括号外的文字

Question

即

$text = 'remove this text (keep this text and 123)';

echo preg_replace('', '', $text);

它应该输出：

(keep this text and 123)

Answer 1

这将做到这一点：（并且也适用于嵌套 () ）

$re = '/[^()]*+(\((?:[^()]++|(?1))*\))[^()]*+/';
$text = preg_replace($re, '$1', $text);

以下是几个测试用例：

Input:
Non-nested case: 'remove1 (keep1) remove2 (keep2) remove3'
Nested case:     'remove1 ((keep1) keep2 (keep3)) remove2'

Output:
Non-nested case: '(keep1)(keep2)'
Nested case:     '(keep1) keep2 (keep3)'

Answer 2

获取括号内找到的任何内容，将其放入捕获组中并仅保留它，如下所示：

echo preg_replace('/^.*(\(.*\)).*$/', '$1', $text);

Answer 3

这里是“非 preg_replace”方式：

<?

$text = 'remove this text (keep this text)' ;

$start = strpos($text,"(") ; 
$end = strpos($text,")") ; 

echo substr($text,$start+1,$end-$start-1) ; // without brackets
echo substr($text,$start,$end-$start+1) ; // brackets included

?>

注意：
- 这仅提取第一对括号。
- 将 strpos() 替换为 strrpos() 以获得最后一对括号。
- 嵌套的括号会引起麻烦。