php - 正则表达式:如果字符串包含括号内的特定单词，则删除括号及其内容

Question

使用正则表达式，我想检测字符串中括号内是否存在特定单词，如果存在，请删除括号及其内容。

我想要定位的词是:

picture
see
lorem

因此，这里有 3 个字符串示例:

$text1 = 'Hello world (see below).';
$text2 = 'Lorem ipsum (there is a picture here) world!';
$text3 = 'Attack on titan (is lorem) great but (should not be removed).';

我可以使用什么正则表达式 preg_replace() :

$text = preg_replace($regex, '' , $text);

删除这些括号及其内容(如果它们包含这些词)？

结果应该是:

$text1 = 'Hello world.';
$text2 = 'Lorem ipsum world!';
$text3 = 'Attack on titan great but (should not be removed).';

这是一个 ideone用于检测。

Answer 1

您可以使用以下方法(感谢@Casimir 之前指出错误!):

<?php
$regex = '~
            (\h*\(                             # capture groups, open brackets
                [^)]*?                         # match everything BUT a closing bracket lazily
                (?i:picture|see|lorem)         # up to one of the words, case insensitive
                [^)]*?                         # same construct as above
            \))                                # up to a closing bracket
            ~x';                               # verbose modifier

$text = array();
$text[] = 'Hello world (see below).';
$text[] = 'Lorem ipsum (there is a picture here) world!';
$text[] = 'Attack on titan (is lorem) great but (should not be removed).';

for ($i=0;$i<count($text);$i++)
    $text[$i] = preg_replace($regex, '', $text[$i]);

print_r($text);
?>

见 a demo on ideone.com和 on regex101.com .