Question

I'm learning Ajax currently, and I use "Sams Teach Yourself Ajax in 10 minutes". I've learnt much, and understand Ajax basically. The book writes code-examples and go through the code bit by bit to explain what each bit does. However. In this code there is something wrong written by the author, and I don't know what..

İşte kod:

<html>
<head>
<title>An Ajax RSS Headline Reader</title>
</head>
<script language="JavaScript" type="text/javascript">
function getXMLHTTPRequest() {
try {
req = new XMLHttpRequest(); /* e.g. Firefox */
} catch(e) {
  try {
  req = new ActiveXObject("Msxml2.XMLHTTP");
  /* some versions IE */
  } catch (e) {
    try {
    req = new ActiveXObject("Microsoft.XMLHTTP");
    /* some versions IE */
    } catch (E) {
      req = false;
    }
  }
}
return req;
}
var http = getXMLHTTPRequest();

function getRSS() {
  var myurl = 'rssproxy.php?feed=';
  var myfeed = document.form1.feed.value;
    myRand = parseInt(Math.random()*999999999999999);
    // cache buster

   var modurl = myurl+escape(myfeed)+"&rand="+myRand;
   http.open("GET", modurl, true);
   http.onreadystatechange = useHttpResponse;
   http.send(null);
}
function useHttpResponse() {
   if (http.readyState == 4) {
    if(http.status == 200) {
       // first remove the childnodes
       // presently in the DM
       while (document.getElementById('news') .hasChildNodes())
      {
document.getElementById('news').removeChild(document .getElementById('news').firstChild);
      }
      var titleNodes = http.responseXML .getElementsByTagName("title");
      var descriptionNodes = http.responseXML .getElementsByTagName("description");
      var linkNodes = http.responseXML .getElementsByTagName("link");
      for(var i =1;i<titleNodes.length;i++)
      {
        var newtext = document .createTextNode(titleNodes[i] .childNodes[0].nodeValue);
        var newpara = document.createElement('p');
        var para = document.getElementById('news') .appendChild(newpara);
        newpara.appendChild(newtext);
        newpara.className = "title";

        var newtext2 = document .createTextNode(descriptionNodes[i] .childNodes[0].nodeValue);
        var newpara2 = document.createElement('p');
        var para2 = document .getElementById('news').appendChild(newpara2);
        newpara2.appendChild(newtext2);
        newpara2.className = "descrip";
        var newtext3 = document .createTextNode(linkNodes[i] .childNodes[0].nodeValue);
        var newpara3 = document.createElement('p');

        var para3 = document.getElementById('news') .appendChild(newpara3);
        newpara3.appendChild(newtext3);
        newpara3.className = "link";
      }
    }
  }
}
</script>
<body>
<center>
<h3>An Ajax RSS Reader</h3>
<form name="form1">
URL of RSS feed: <input type="text" name="feed" size="50" value="http://"><input
 type="button" onClick="getRSS()" value="Get Feed"><br><br>
<div id="news" class="displaybox"> <h4>Feed Titles</h4></div>
</form>
</center>
</html>

ve rssproxy.php bu gibi görünüyor:

<?php
$mysession = curl_init($_GET['feed']);
curl_setopt($mysession, CURLOPT_HEADER, false);
curl_setopt($mysession, CURLOPT_RETURNTRANSFER, true);
$out = curl_exec($mysession);
header("Content-Type: text/xml");
echo $out;
curl_close($mysession);
?>

Bana yardımcı olabilir, ben bunu takdir ediyorum!

Tam (elbette arıza olmadan) aynı kodları sürece ben sadece, bu kod ile yanlış ne olduğunu bilmek istiyorum, ve diğer Ajax RSS-öğreticiler bağlantıları alamadım ..

Answer 1

Öncelikle php script hiç bir rss feed getirme ve teslim bulamayacağını test edelim

<?php
error_reporting(E_ALL); ini_set('display_errors', 1);
// $mysession = curl_init($_GET['feed']);
$mysession = curl_init('http://www.spiegel.de/schlagzeilen/index.rss');


curl_setopt($mysession, CURLOPT_HEADER, false);
curl_setopt($mysession, CURLOPT_RETURNTRANSFER, true);
$out = curl_exec($mysession);
header("Content-Type: text/xml");
echo $out;
curl_close($mysession);
?>

Just invoke it from the browser and check the output.
If that works install and use a client-side script debugger. E.g. for firefox use Firebug. Javascript errors will show up in the console tab.

Answer 2

Ben XML içeriğinin yanlış bir şey düşünüyorum. Belki <link> etiketi eksik.

Ajax RSS-okuyucu: kodu ile Yardım, eksik / yanlış ne?

2 Cevap

etiketler