我的Facebook页面Feed看起来像这样,
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 ......... featureog(仅提取图片网址)之间的内容? 感谢任何代码示例。
答案 0 :(得分:1)
我的正则表达式很丑陋,但却有诀窍:
<?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];
?>
答案 1 :(得分:1)
$parts = parse_url('http://external.ak.fbcdn.net/safe_image.php?d=AQA-U_vFlmf0YW5c&w=130&h=130&url=http%3A%2F%2Fi1.ytimg.com%2Fvi%2FX9Hx6nUTSwE%2Fmaxresdefault.jpg%3Ffeature%3Dog');
parse_str($parts['query'], $params);
var_dump($params['url']);