Home >php教程 >php手册 >php获取301跳转URL简单实例

php获取301跳转URL简单实例

WBOY
WBOYOriginal
2016-06-06 20:26:061132browse

这篇文章主要介绍了php获取301跳转URL简单实例,有需要的朋友可以参考一下

复制代码 代码如下:


/**
 * get_redirect_url()
 * Gets the address that the provided URL redirects to,
 * or FALSE if there's no redirect.
 *
 * @param string $url
 * @return string
 */
function get_redirect_url($url){
    $redirect_url = null;

    $url_parts = @parse_url($url);
    if (!$url_parts) return false;
    if (!isset($url_parts['host'])) return false; //can't process relative URLs
    if (!isset($url_parts['path'])) $url_parts['path'] = '/';

    $sock = fsockopen($url_parts['host'], (isset($url_parts['port']) ? (int)$url_parts['port'] : 80), $errno, $errstr, 30);
    if (!$sock) return false;

    $request = "HEAD " . $url_parts['path'] . (isset($url_parts['query']) ? '?'.$url_parts['query'] : '') . " HTTP/1.1\r\n";
    $request .= 'Host: ' . $url_parts['host'] . "\r\n";
    $request .= "Connection: Close\r\n\r\n";
    fwrite($sock, $request);
    $response = '';
    while(!feof($sock)) $response .= fread($sock, 8192);
    fclose($sock);
    if (preg_match('/^Location: (.+?)$/m', $response, $matches)){
        if ( substr($matches[1], 0, 1) == "/" )
            return $url_parts['scheme'] . "://" . $url_parts['host'] . trim($matches[1]);
        else
            return trim($matches[1]);

    } else {
        return false;
    }

}
/**
 * get_all_redirects()
 * Follows and collects all redirects, in order, for the given URL.
 *
 * @param string $url
 * @return array
 */
function get_all_redirects($url){
    $redirects = array();
    while ($newurl = get_redirect_url($url)){
        if (in_array($newurl, $redirects)){
            break;
        }
        $redirects[] = $newurl;
        $url = $newurl;
    }
    return $redirects;
}


php实现用socket获取301跳转地址,,可以提取跳转过程中的url
Statement:
The content of this article is voluntarily contributed by netizens, and the copyright belongs to the original author. This site does not assume corresponding legal responsibility. If you find any content suspected of plagiarism or infringement, please contact admin@php.cn