PHP regex - 用链接替换多个标签之间的文本

debugcn 发表于 Dev

马克西

我有看起来像这样的字符串

A <<double>> <<tripod>> (for holding a <<plate>>, etc.) with six feet, of which three rest on the ground, in whatever position it is placed.

我怎样才能以<<>>之间的所有内容成为链接的格式替换它。

结果应如下所示：

A <a href="search/double">double</a> <a href="search/tripod">tripod</a> (for holding a <a href="search/plate">plate</a>, etc.) with six feet, of which three rest on the ground, in whatever position it is placed.

迈克尔·K

这样的事情应该工作：

$str = "A <<double>> <<tripod>> (for holding a <<plate>>, etc.) with six feet, of which three rest on the ground, in whatever position it is placed.";

    preg_match_all("|<<(.*)>>|U", $str, $out, PREG_PATTERN_ORDER);

    for($i=0; $i < count($out[0]); $i++) {
        $str = str_replace($out[0][$i], '<a href="search/'. $out[1][$i] .'">'. $out[1][$i] .'</a>', $str);
    }

echo $str;

另一种选择是使用preg_replace_callback并创建一个递归函数：

function parseTags($input) {
    $regex = "|<<(.*)>>|U";

    if (is_array($input)) {
        $input = '<a href="search/'.$input[1].'">'.$input[1].'</a>';
    }

return preg_replace_callback($regex, 'parseTags', $input);
}

$str = "A <<double>> <<tripod>> (for holding a <<plate>>, etc.) with six feet, of which three rest on the ground, in whatever position it is placed.";

echo parseTags($str);

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