PHP獲取URL一個字符串和一些更

Question

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

字符串而現在，我需要做下面的事情。

1. 從上面的字符串http://www.anydomain.com/pop_new.php?sid=10623&aid=1581&rand=0.6808111508818073
2. 獲取替換URL網址{％URL％}所以它應該看起來像The theme song of whatever - {%url%} #string

目前我使用下面的代碼，但它無法取代上面的url。

$urlregex_ = "(https?)\:\/\/[a-z0-9+\$_-]+(\.[a-z0-9+\$_-]+)*(\/([a-z0-9+\$_-]\.?)+)*\/?(\?[a-z+&\$_.-][a-z0-9;:@/&%=+\$_.-]*)?(#[a-z_.-][a-z0-9+\$_.-]*)?"; 
preg_match('~'.$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);` 

希望能幫到你，謝謝你，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"); 
    }