Question

我在扩展某些代码时遇到了问题。

我基本上需要在现有的xml字段中添加一个新的xml字段。

以下是完整代码：

HTML文件

<html>
<head>
<script type="text/javascript">
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="MSNBC">MSNBC News</option>
</select>
</form>
<br />
<div id="rssOutput">RSS-feed will be listed here...</div>
</body>
</html>

PHP文件

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

//find out which feed was selected
if($q=="Google")
  {
  $xml=("http://feeds.bbci.co.uk/news/rss.xml");
  }
elseif($q=="MSNBC")
  {
  $xml=("http://feeds.bbci.co.uk/news/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>");
  }
?>

由于

Answer 1

经过大量研究后，我得出的结论是，最好的方法是使用SimpleXML。

对于有同样问题的人，这里是指向SimpleXML文档的链接：

http://php.net/manual/en/book.simplexml.php

Answer 2

$item_thumb=$x->item($i)->getElementsByTagName('thumbnail')->item(0)->childNodes->item(0)->nodeValue;

添加新的XML标记代码（PHP，HTML）

2 个答案: