我有一个像这样的Facebook页面Feed，
http://external.ak.fbcdn.net/safe_image.php?d=AQA-U_vFlmf0YW5c&w=130&h=130&url=http%3A%2F%2Fi1.ytimg.com%2Fvi%2FX9Hx6nUTSwE%2Fmaxresdefault.jpg%3Ffeature%3Dog
如何提取&url之间的内容.........featuredog(只提取图像url) ?感谢任何代码示例。
我的正则表达式是丑陋的，但做的伎俩:
<?php
$str = 'http://external.ak.fbcdn.net/safe_image.php?d=AQA-U_vFlmf0YW5c&w=130&h=130&url=http%3A%2F%2Fi1.ytimg.com%2Fvi%2FX9Hx6nUTSwE%2Fmaxresdefault.jpg%3Ffeature%3Dog';
$str = urldecode($str);
preg_match_all('~&url=(.*?)['?'!]?feature~i', $str, $matches, PREG_PATTERN_ORDER);
echo $matches[1][0];
?>
$parts = parse_url('http://external.ak.fbcdn.net/safe_image.php?d=AQA-U_vFlmf0YW5c&w=130&h=130&url=http%3A%2F%2Fi1.ytimg.com%2Fvi%2FX9Hx6nUTSwE%2Fmaxresdefault.jpg%3Ffeature%3Dog');
parse_str($parts['query'], $params);
var_dump($params['url']);