删除括号外的文本

Question

即

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

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

应输出：

(keep this text and 123)

Answer 1

这样做:(并与嵌套的()一起使用）

$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

在括号内找到任何内容，将其放入捕获组并仅保留，如下所示：

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

Answer 3

这里是'non 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：
- 这只提取第一对括号 - 用strrpos（）替换strpos（）以获得最后一对括号 - 嵌套括号会造成麻烦。