php – 使用Phonegap应用程序执行ajax请求的问题

当我在浏览器中试用代码时,我设法让这个工作得很好. php文件获取feed并输出它就像我期望的那样.但是当我从我编译的Phonegap应用程序中运行相同的文件时,ajax-request只返回php文件的内容(php代码,而不是执行结果).

我花了几个小时谷歌搜索这个并尝试了很多教程和调整.我在官方的Phonegap论坛上找不到任何解决方案.我究竟做错了什么？问题似乎是PHP没有响应请求.我试图将php文件移动到另一个域,但结果是一样的,它在我的浏览器中工作,但不在编译的应用程序中.

这是启动ajax代码的jQuery代码：

function get_rss_feed() {
    //clear the content in the div for the next feed.
    $("#feed_content").empty().html('<img class="loader" src="js/images/ajax-loader.gif" alt=""/>');

    $.ajax({
        url: 'http://192.168.1.7/rssApp/www/rss-proxy.php?url=http://www.nytimes.com/services/xml/rss/nyt/GlobalHome.xml',success: function parseRSS(d) {

        //find each 'item' in the file and parse it
        $(d).find('item').each(function() {

            //name the current found item this for this particular loop run
            var $item = $(this);
            // grab the post title
            var title = $item.find('title').text();
            // grab the post's URL
            var link = $item.find('link').text();
            // next,the description
            var description = $item.find('description').text();
            //don't forget the pubdate
            var pubDate = $item.find('pubDate').text();

            // now create a var 'html' to store the markup we're using to output the feed to the browser window
            var html = "<div class="entry"><h2 class="postTitle">" + title + "</h2>";
            html += "<em class="date">" + pubDate + "</em>";
            html += "<p class="description">" + description + "</p>";
            html += "<a href="" + link + "" target="_blank">Read More >></a></div>";

            //put that feed content on the screen!
            $('#feed_content').append($(html));
        });
        $('#feed_content img.loader').fadeOut();
    }

    });

};

这是rss-proxy.php,它从url加载XML并输出：

<?php
    // PHP Proxy
    // Loads a XML from any location. Used with Flash/Flex apps to bypass security restrictions
    // Author: Paulo Fierro
    // January 29,2006
    // usage: proxy.php?url=http://mysite.com/myxml.xml

    $session = curl_init($_GET['url']);                    // Open the Curl session
    curl_setopt($session,CURLOPT_HEADER,false);          // Don't return HTTP headers
    curl_setopt($session,CURLOPT_RETURNTRANSFER,true);   // Do return the contents of the call
    $xml = curl_exec($session);                            // Make the call
    header("Content-Type: text/xml");                  // Set the content type appropriately
    echo $xml;        // Spit out the xml
    curl_close($session); // And close the session
?>