Jump to content

error when trying to paste url into ajax function


becu

Recommended Posts

Hi guy,

I'm very new to ajax. Please help me fix error below.

 

This code supposes to take the url from the link, put it into showHint function and output and image.

 

<script type="text/javascript">
var xmlHttp

function showHint(str)
{
if (str.length==0)
{ 
document.getElementById("txtHint").innerHTML="";
return;
}
xmlHttp=GetXmlHttpObject()
if (xmlHttp==null)
{
alert ("Your browser does not support AJAX!");
return;
} 
var url="img.php";
url=url+"?img="+str;
url=url+"&sid="+Math.random();
xmlHttp.onreadystatechange=stateChanged;
xmlHttp.open("GET",url,true);
xmlHttp.send(null);
} 

function stateChanged() 
{ 
if (xmlHttp.readyState==4)
{ 
document.getElementById("txtHint").innerHTML=xmlHttp.responseText;
}
}

function GetXmlHttpObject()
{
var xmlHttp=null;
try
{
// Firefox, Opera 8.0+, Safari
xmlHttp=new XMLHttpRequest();
}
catch (e)
{
// Internet Explorer
try
{
xmlHttp=new ActiveXObject("Msxml2.XMLHTTP");
}
catch (e)
{
xmlHttp=new ActiveXObject("Microsoft.XMLHTTP");
}
}
return xmlHttp;
}
</script>
<a href="#" onclick="showHint(http://farm2.static.flickr.com/1123/847641124_05ad612bf7.jpg)">here</a>
<span id="txtHint"></span>

 

And this is the code of img.php

<?php
echo "<img src=\"{$_GET['img']}\">";
?>

 

the code won't work. when I paste the source code of the page into frontpage, it said I got an error near onclick. When I replace the link of the image by a random number or any other character, it works just fine. why?

 

thanks.

Archived

This topic is now archived and is closed to further replies.

×
×
  • Create New...

Important Information

We have placed cookies on your device to help make this website better. You can adjust your cookie settings, otherwise we'll assume you're okay to continue.