ansaurus

Question

Answer 1

+1 A:

I would verify that the status is what you expect by adding a bit of debug output. If it is returning status 0, you probably have a cross-domain security issue.

<script type="text/javascript">
var t;
function tick() {
  xmlhttp=new XMLHttpRequest();
  xmlhttp.onreadystatechange=function() {
      document.getElementById("debug").innerHTML = "readyState: " + xmlhttp.readyState + "<br />status: " + xmlhttp.status;
      if (xmlhttp.readyState==4 && xmlhttp.status==200 && document.getElementById("txtHint").innerHTML!=xmlhttp.responseText) {
            document.getElementById("txtHint").innerHTML=xmlhttp.responseText;
        }
      };
  xmlhttp.open("GET","http://ubarskit.com/v/response.php",true);
  xmlhttp.send();
  t=setTimeout("tick()",1000);
}
</script>
</head>
<body onload="tick()"> 
<span id="debug"></span>
<span id="txtHint"></span>
</body>

nFreeze 2010-10-25 23:05:38

Thanks, you where right about the cross-domain issue, changed the address to "respose.php" and things started working, i Firefox 3.6 at least :S

Ahtenus 2010-10-26 15:06:43

Answer 2

A:

Instead of doing it like real men, I used jQuery ;)

<script type="text/javascript">
function tick() {
$.get('response.php',function(data) {
  $('#txtHint').html(data);
});
}
function submitShout(form){
$.get('response.php',{ ropa: form.elements['ropa'].value});
}
</script>
</head>
<body onload="window.setInterval('tick()',1000)"> 
<span id="txtHint"></span>
<form action="response.php" method="get" onsubmit="submitShout(this); return false;">
  <input type="text" id="ropa" name="ropa">
</form>

Ahtenus 2010-10-26 15:14:57

ansaurus

tags:

views:

answers:

Whats wrong with my ajax request?

related questions