根据@JoyceBabu在这篇文章Get http-statuscode without body using cURL?应该可以通过fsockopen获取一个URL的http状态码。阅读http状态码fsockopen
所以我从@JoyceBabu其工作代码:
<?php
$fp = fsockopen("www.google.com", 80, $errno, $errstr, 30);
if ($fp) {
$out = "GET/HTTP/1.1\r\n";
$out .= "Host: www.google.com\r\n";
$out .= "Accept-Encoding: gzip, deflate, sdch\r\n";
$out .= "Accept-Language: en-GB,en-US;q=0.8,en;q=0.6\r\n";
$out .= "User-Agent: Mozilla/5.0 (Macintosh; Intel Mac OS X 10_7_5) AppleWebKit/537.36 (KHTML, like Gecko) Chrome/39.0.2171.71 Safari/537.36\r\n";
$out .= "Accept: text/html,application/xhtml+xml,application/xml;q=0.9,image/webp,*/*;q=0.8\r\n";
$out .= "Connection: Close\r\n\r\n";
fwrite($fp, $out);
$tmp = explode(' ', fgets($fp, 13));
echo $tmp[1];
fclose($fp);
}
然后,我改变了URL读给www.raffiniert.biz/aktuell:
<?php
$fp = fsockopen("www.raffiniert.biz", 80, $errno, $errstr, 30);
if ($fp) {
$out = "GET/HTTP/1.1\r\n";
$out .= "Host: www.raffiniert.biz/aktuell\r\n";
$out .= "Accept-Encoding: gzip, deflate, sdch\r\n";
$out .= "Accept-Language: en-GB,en-US;q=0.8,en;q=0.6\r\n";
$out .= "User-Agent: Mozilla/5.0 (Macintosh; Intel Mac OS X 10_7_5) AppleWebKit/537.36 (KHTML, like Gecko) Chrome/39.0.2171.71 Safari/537.36\r\n";
$out .= "Accept: text/html,application/xhtml+xml,application/xml;q=0.9,image/webp,*/*;q=0.8\r\n";
$out .= "Connection: Close\r\n\r\n";
fwrite($fp, $out);
$tmp = explode(' ', fgets($fp, 13));
echo $tmp[1];
fclose($fp);
}
返回HTTP 400 - 这是不正确的。
又如:www.raffiniert.biz/kunden回报301这应该是403
任何想法我做错了吗?
感谢 拉斐尔
我看到,GET线在我的代码是“/”...我累了,对不起:)无论如何,它不适用于某些网址,我收集了一些示例 – 2014-12-01 21:52:33
@RaphaelJeger查看更新的答案 – outlyer 2014-12-01 22:53:08
谢谢,我在此期间看到了。现在这是一个快速解决方案,但它应该可以解决最终的重定向... – 2014-12-01 23:02:15