使用 PHP 从 RSS 解析 XML 并使用 AJAX 传递

Question

我想在 rss link（路透社：头条新闻）中获取最新的 ITEM，当新的 ITEM 出现时，PHP/AJAX 会将旧的 ITEM 替换为新的.

我尝试使用 PHP 和 AJAX 来做到这一点，但我得到的是空白页或 "Fatal error: Maximum execution time of 30 seconds exceeded in C:\xampp\htdocs\test4.php on line 11"。

函数：

   <?php function parserSide($feedURL) { $rss = simplexml_load_file($feedURL); $output = "
<ul class='newsSide'>"; $i = 0; foreach ($rss->channel->item as $feedItem) { $i++; $output .= "
    <li><a href='$feedItem->link' title='$feedItem->title'>" . $feedItem->title . "</a></li>"; if($i >= 1) break; } $output .= "</ul>"; return $output; } ?>
<?php
    if(isset($_GET['fetchOnly'])){
        $url = 'http://feeds.reuters.com/reuters/topNews'; 

        $response = array();
        $handle = fopen($url, 'r'); 
        if ($handle) { 
            while (($data = simplexml_load_file($url)) !== FALSE) 
            { 
                $response = parserSide($url);                 
            }
            fclose($handle);
        }  
        echo $response;
        die();
    } 
?>
    <div id="responseText"></div>
    <script>
        // run the function, it will re-run itself
        fetchRate();

        function fetchRate() {
            // create the new AJAX Object
            xmlhttp = new XMLHttpRequest();
            // this handles the request
            xmlhttp.onreadystatechange = function () {
                if (xmlhttp.readyState == XMLHttpRequest.DONE) {
                    // if the request came back successfully
                    if (xmlhttp.status == 200) {
                        // write the response to a div
                        div = document.getElementById("responseText")
                        div.innerHTML = xmlhttp.responseText;
                    } else {
                        // if the request had an error
                        div.innerHTML = div.innerHTML + '<br />Error fetching rates error code : ' + xmlhttp.status;
                    }
                    // rerun this function to fetch updates
                    setTimeout(fetchRate, 1000);
                }
            };
            // open the AJAX Object
            xmlhttp.open("GET", "<?= basename(__FILE__) ?>?fetchOnly", true);
            // send the AJAX request
            xmlhttp.send();
        }
    </script>

提前致谢

Answer 1

如果加载失败，

简单 xml 将 return false，因此只需将其用于检查。单点可靠校验。

<?php
    function parserSide($feedURL) {
    $rss = simplexml_load_file($feedURL);
    $output = "";
    if($rss){
        //simple_xml load was a success. It does everything
        $output .= "<ul class='newsSide'>";
        $i = 0;
        foreach ($rss->channel->item as $feedItem) {
            $i++;
            $output .= "<li><a href='$feedItem->link' title='$feedItem->title'>" . $feedItem->title . "</a></li>";
            if($i >= 1) break;
        }
        $output .= "</ul>";
    }
    return $output;
}
if(isset($_GET['fetchOnly'])){
    $url = 'http://feeds.reuters.com/reuters/topNews';
    $handle = fopen($url, 'r');
    $response = parserSide($url);
    echo $response;
    die();
}
?>
<div id="responseText"></div>
<script>
    // run the function, it will re-run itself
    fetchRate();
    function fetchRate() {
        var div = document.getElementById("responseText")
        div.innerHTML = "attempting to load data from server....";
        // create the new AJAX Object
        xmlhttp = new XMLHttpRequest();
        // this handles the request
        xmlhttp.onreadystatechange = function () {
            if (xmlhttp.readyState == XMLHttpRequest.DONE) {
                // if the request came back successfully
                if (xmlhttp.status == 200) {
                    // write the response to a div
                    div.innerHTML = xmlhttp.responseText;
                } else {
                    // if the request had an error
                    div.innerHTML = div.innerHTML + '<br />Error fetching rates error code : ' + xmlhttp.status;
                }
                // rerun this function to fetch updates
                setTimeout(fetchRate, 1000);
            }
        };
        // open the AJAX Object
        xmlhttp.open("GET", "?fetchOnly", true);
        // send the AJAX request
        xmlhttp.send();
    }
</script>

使用 PHP 从 RSS 解析 XML 并使用 AJAX 传递

Parse XML from RSS with PHP and pass using AJAX

javascript

php

xml

ajax

rss