PHP 转义字符后带有单词边界的正则表达式

Question

我最近无意中发现了这个问题，但我不明白为什么会这样。

考虑以下示例：我有一个随机文本和一个包含一些编程语言的数组。在一个循环中，我将使用正则表达式和单词边界 \b 前后匹配每种语言作为整个单词，然后打印 URL.

$string = 'I don\'t know C e C++ so well, but I can code in PHP.';
$languages = [
    'PHP' => '/php/',
    'C++' => '/cpp/',
    'C' => '/c/',
];

foreach ($languages as $name => $uri) {
    $regex = '/\b' . preg_quote($name, '/') . '\b/';
    if (preg_match($regex, $string)) {
        echo "For {$name} information refer to http://foo.bar{$uri}" . PHP_EOL;
    }
}

我希望得到以下输出：

For PHP information refer to http://foo.bar/php/
For C++ information refer to http://foo.bar/cpp/
For C information refer to http://foo.bar/c/

然而，我得到的输出是：

For PHP information refer to http://foo.bar/php/
For C information refer to http://foo.bar/c/

紧跟转义加号 (+) 之后的单词边界 (\b) 无法正常工作。

如果我将 \b 替换为 [^\w] 它会起作用，但我不能 100% 确定这种方法不会适得其反。

为什么会发生这种情况，如何才能获得我需要的结果？

Answer 1

解决此问题的推荐方法是使用环视来断言单词字符而不是边界，例如(?<!\w)c\+\+(?!\w):

$string = 'I don\'t know C e C++ so well, but I can code in PHP.';
$languages = [
    'PHP' => '/php/',
    'C++' => '/cpp/',
    'C' => '/c/',
];

foreach ($languages as $name => $uri) {
    $regex = '/(?<!\w)' . preg_quote($name, '/') . '(?!\w)/';
    if (preg_match($regex, $string)) {
        echo "For {$name} information refer to http://foo.bar{$uri}" . PHP_EOL;
    }
}

输出：

For PHP information refer to http://foo.bar/php/
For C++ information refer to http://foo.bar/cpp/
For C information refer to http://foo.bar/c/

PHP 转义字符后带有单词边界的正则表达式

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