PHP AJAX RSS 온로드

-1

현재 w3schools의 RSS 피드 리더 인 W3schools RSS Reader을 사용하고 있습니다.PHP AJAX RSS 온로드

<html> 
<head> 
<script> 
    function showRSS(str) { 
    if (str.length==0) { 
    document.getElementById("rssOutput").innerHTML=""; 
    return; 
    } 
    if (window.XMLHttpRequest) { 
    // code for IE7+, Firefox, Chrome, Opera, Safari 
    xmlhttp=new XMLHttpRequest(); 
    } else { // code for IE6, IE5 
    xmlhttp=new ActiveXObject("Microsoft.XMLHTTP"); 
    } 
    xmlhttp.onreadystatechange=function() { 
    if (xmlhttp.readyState==4 && xmlhttp.status==200) { 
    document.getElementById("rssOutput").innerHTML=xmlhttp.responseText; 
     } 
    } 
    xmlhttp.open("GET","getrss.php?q="+str,true); 
    xmlhttp.send(); 
    } 
</script> 
</head> 
<body> 

<form> 
<select onchange="showRSS(this.value)"> 
<option value="">Select an RSS-feed:</option> 
<option value="Google">Google News</option> 
<option value="NBC">NBC News</option> 
</select> 
</form> 
<br> 
<div id="rssOutput">RSS-feed will be listed here...</div> 
</body> 
</html>

과 PHP 코드 : 나는로드 구글 뉴스 옵션을 얻는 방법

을 : 나는 시도

<?php 
//get the q parameter from URL 
$q=$_GET["q"]; 

//find out which feed was selected 
if($q=="Google") { 
    $xml=("http://news.google.com/news?ned=us&topic=h&output=rss"); 
} elseif($q=="NBC") { 
    $xml=("http://rss.msnbc.msn.com/id/3032091/device/rss/rss.xml"); 
} 

$xmlDoc = new DOMDocument(); 
$xmlDoc->load($xml); 

//get elements from "<channel>" 
$channel=$xmlDoc->getElementsByTagName('channel')->item(0); 
$channel_title = $channel->getElementsByTagName('title') 
->item(0)->childNodes->item(0)->nodeValue; 
$channel_link = $channel->getElementsByTagName('link') 
->item(0)->childNodes->item(0)->nodeValue; 
$channel_desc = $channel->getElementsByTagName('description') 
->item(0)->childNodes->item(0)->nodeValue; 

//output elements from "<channel>" 
echo("<p><a href='" . $channel_link 
    . "'>" . $channel_title . "</a>"); 
echo("<br>"); 
echo($channel_desc . "</p>"); 

//get and output "<item>" elements 
$x=$xmlDoc->getElementsByTagName('item'); 
for ($i=0; $i<=2; $i++) { 
    $item_title=$x->item($i)->getElementsByTagName('title') 
    ->item(0)->childNodes->item(0)->nodeValue; 
    $item_link=$x->item($i)->getElementsByTagName('link') 
    ->item(0)->childNodes->item(0)->nodeValue; 
    $item_desc=$x->item($i)->getElementsByTagName('description') 
    ->item(0)->childNodes->item(0)->nodeValue; 
    echo ("<p><a href='" . $item_link 
    . "'>" . $item_title . "</a>"); 
    echo ("<br>"); 
    echo ($item_desc . "</p>"); 
} 
?>

변경 여기

는 HTML 코드입니다 처음 웹 사이트를 선택하지 않고 웹 사이트가 열릴 때 자동으로?

출처

2014-09-15 Codelove

PHP의 simplexml_load_file을 사용하는 훨씬 간단한 해결책을 발견했습니다. 전체 코드는 다음과 같습니다.

<!-- language: lang-php --> 
<?php 
$html = ""; 
$url = "http://news.google.com/news?ned=us&topic=h&output=rss"; 
$xml = simplexml_load_file($url); 
for($i = 0; $i < 10; $i++){ 
$title = $xml->channel->item[$i]->title; 
$link = $xml->channel->item[$i]->link; 
$description = $xml->channel->item[$i]->description; 
$pubDate = $xml->channel->item[$i]->pubDate; 
$html .= "<a href='$link'><h3>$title</h3></a>"; 
$html .= "$description"; 
$html .= "<br />$pubDate"; 
} 
echo $html; 
?>

출처

2014-09-26 05:25:08 Codelove

PHP AJAX RSS 온로드

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