Question

我有一个像The theme song of whatever - http://www.anydomain.com/pop_new.php?sid=10623&aid=1581&rand=0.6808111508818073 #string

这样的字符串

现在，我需要做以下事情。

从字符串http://www.anydomain.com/pop_new.php?sid=10623&aid=1581&rand=0.6808111508818073
将网址替换为{％url％}，以使它看起来像The theme song of whatever - {%url%} #string

目前我使用的是以下代码，但无法替换上述网址。

$urlregex_ = "(https?)\:\/\/[a-z0-9+\$_-]+(\.[a-z0-9+\$_-]+)*(\/([a-z0-9+\$_-]\.?)+)*\/?(\?[a-z+&\$_.-][a-z0-9;:@/&%=+\$_.-]*)?(#[a-z_.-][a-z0-9+\$_.-]*)?";
preg_match('~'.$urlregex_.'~',preg_replace('/\+/',' ',$url),$url_only);
$url_ = preg_replace('/ /','+',$url_only[0]);
$text = preg_replace('~'.$url_.'~','{%url%} ',$url);
return array('url' => $url_only[0], 'text' => $text);`

希望你能帮忙，谢谢，pnm123

Answer 1

<?php
    $orig = "The theme song of whatever - http://www.anydomain.com/pop_new.php" .
        "?sid=10623&aid=1581&rand=0.6808111508818073 #string";
    $urlregex = '~(?:https?)://[a-z0-9+$_-]+(?:\\.[a-z0-9+$_-]+)*' .
        '(?:/(?:[a-z0-9+$_-]\\.?)+)*/?(?:\\?[a-z+&$_.-][a-z0-9;:@/&%=+$_.-]*)?'.
        '(?:#[a-z_.-][a-z0-9+$_.-]*)?~i';
    if (preg_match($urlregex, $orig, $matches)) {
        $after = preg_replace($urlregex, "{%url%}", $orig);
        var_dump(array('url' => $matches[0], 'text' => $after));
    } else { //no array found
        die("oops");
    }

PHP从字符串中获取url等等

1 个答案: