如何在不离开网站的情况下显示链接中的信息？ PHP

Question

我得到的网站完全正常工作，并从rss饲料返回信息，因为它返回与网站链接我如何显示信息从链接而不是重定向用户回到原来的网站？

<?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

使用卷曲从链接读取信息并发布它，例如，

$ch = curl_init(); 
curl_setopt($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); 

没有必要。你可以在你的prog中添加这个代码

<?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 - '; 

} 

?>