如何在不离开网站的情况下显示链接中的信息？ PHP

Question

I got the site fully working and returning the information from the rss feed since the it return with website link how can i display the information from the link instead of redirect the user back to the original site?

<?php

    $query = 'http://query.yahooapis.com/v1/public/yql?q=Select%20*%20from%20rss%20where%20url%3D%22http%3A%2F%2Fnestor.sunderland.ac.uk%2F~bf71wx%2FphpTest%2Fjquery.mobile-1.0.1%2520-%2520%25E8%25A4%2587%25E8%25A3%25BD%2Fjquery.mobile-1.0.1%2Fdemos%2Fdocs%2Frss%2FNews.xml%22%20&diagnostics=true';
    $xml = simplexml_load_file($query);
    //var_dump($xml);

    echo '<h2>World of Tank News</h2>';

    //iterate over query result set
    $results = $xml->results;
    foreach ($results->item as $r){
        echo $r->title . "<br />";
        echo "<a href=\"" . $r->link . "\">" . $r->link . "</a><br /><br />";
        echo 'Publish Date - ';

    }

?>

Answer 1

use curl to read the info from link and publish it e.g.

$ch = curl_init();
curl_setopt($ch, CURLOPT_URL, $file);
curl_setopt($ch, CURLOPT_RETURNTRANSFER, 1);
curl_setopt($ch, CURLOPT_REFERER, $_SERVER['REQUEST_URI']);
$result = curl_exec($ch);
curl_close($ch);

no need. you can add this code in your prog e.g.

<?php

$query = 'http://query.yahooapis.com/v1/public/yql?q=Select%20*%20from%20rss%20where%20url%3D%22http%3A%2F%2Fnestor.sunderland.ac.uk%2F~bf71wx%2FphpTest%2Fjquery.mobile-1.0.1%2520-%2520%25E8%25A4%2587%25E8%25A3%25BD%2Fjquery.mobile-1.0.1%2Fdemos%2Fdocs%2Frss%2FNews.xml%22%20&diagnostics=true';
$xml = simplexml_load_file($query);
//var_dump($xml);

echo '<h2>World of Tank News</h2>';

//iterate over query result set
$results = $xml->results;
foreach ($results->item as $r){
echo $r->title . "<br />";
$ch = curl_init();

$timeout = 5;
curl_setopt($ch,CURLOPT_URL,$r->link);
curl_setopt($ch,CURLOPT_RETURNTRANSFER,1);
curl_setopt($ch,CURLOPT_CONNECTTIMEOUT,$timeout);
$result = curl_exec($ch);
curl_close($ch);
echo $result;
//        echo "<a href=\"" . $r->link . "\">" . $r->link . "</a><br /><br />";
//        echo 'Publish Date - ';

}

?>