escaped_fragment 请求不解析 php？

发布于 2024-11-16 01:44:15 字数 869 浏览 7 评论 0原文

所以最近我按照 Google 的 AJAX 抓取方法重建了我的网站。

一切工作正常， ?_escaped_fragment_= 请求工作正常，但我注意到页面上的任何 php 都没有被解析，并被注释掉，如下所示：

有谁知道为什么会发生这种情况？这是我索引中的一些 PHP：

$fragment = $_GET['_escaped_fragment_'];
    $file = '' . (isset($fragment) && $fragment != '' && $fragment != '/' ? preg_replace('/\//', '', $fragment) : 'home') . '.html';
    $re = '/(^<[^>]*>)|(\n|\r\n|\t|\s{2,4})*/';

    $handle = fopen($file, 'r');
    if ($handle != false) {
        $content = preg_replace($re, '', fread($handle, filesize($file)));
        fclose($handle);
    } else {
        $content = 'Page not found!';
        header(php_sapi_name() == 'cgi' ? 'Status: 404' : 'HTTP/1.1 404');
    }

我注意到其中有一个正则表达式。也许这有什么关系？我无法完全理解这里发生的事情，我不是专家。如果这会一直这样的话，我不想恢复我的网站..

请帮我一把

原文

so recently ive rebuilt my site following Google's AJAX Crawling methods.

Everything works fine and an ?_escaped_fragment_= request works fine but i've noticed any php on the page is not parsed and is commented out like so: 

Does anyone know why this may be happening?? here is some PHP from my index:

$fragment = $_GET['_escaped_fragment_'];
    $file = '' . (isset($fragment) && $fragment != '' && $fragment != '/' ? preg_replace('/\//', '', $fragment) : 'home') . '.html';
    $re = '/(^<[^>]*>)|(\n|\r\n|\t|\s{2,4})*/';

    $handle = fopen($file, 'r');
    if ($handle != false) {
        $content = preg_replace($re, '', fread($handle, filesize($file)));
        fclose($handle);
    } else {
        $content = 'Page not found!';
        header(php_sapi_name() == 'cgi' ? 'Status: 404' : 'HTTP/1.1 404');
    }

I noticed a regexp in there. Maybe that has something to do with it? I cant fully understand whats going on here, im no expert.
Dont exactly want to put my site back up if this is gonna stay like this..

Please lend me a hand here

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