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07.05.2008 at 04:39AM PDT, ID: 23540530
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8.4

Populating the drop down list using ajax.

Asked by cyberwebservice in Asynchronous Javascript and XML (AJAX), PHP Scripting Language

I am trying to populate a drop down list from database, based on the selection in the first drop down list. I want to do this using AJAX.

When I use the following code, the second drop down list gets populated with database values only if the I select "Home Furnishing Fabric". It displays an empty drop down list for the other selections. I guess the selected value is not passed in the query string properly.

I tried changing the order of option values, but still it displays only for the "Home Furnishing Fabric".
I need help with this.

Thanks.Start Free Trial 
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<html>
<body>
 
<script type="text/javascript">
function ajaxFunction(strurl)
{
var xmlHttp;
try
  {
  // Firefox, Opera 8.0+, Safari
  xmlHttp=new XMLHttpRequest();
  }
catch (e)
  {
  // Internet Explorer
  try
    {
    xmlHttp=new ActiveXObject("Msxml2.XMLHTTP");
    }
  catch (e)
    {
    try
      {
      xmlHttp=new ActiveXObject("Microsoft.XMLHTTP");
      }
    catch (e)
      {
      alert("Your browser does not support AJAX!");
      return false;
      }
    }
  }
  xmlHttp.onreadystatechange=function()
    {
    if(xmlHttp.readyState==4)
      {
      document.getElementById('weavediv').innerHTML=xmlHttp.responseText;
      }
    }
  xmlHttp.open("GET",strurl,true);
  xmlHttp.send(null);
  }
</script>
 
<form name="myForm">
<select name="fablist" onChange="ajaxFunction('test.php?selfab='+this.options[this.selectedIndex].value);">
<option value"Apparel Fabric">Apparel Fabric</option>
<option value"Cotton Spandax Fabric">Cotton Spandax Fabric</option>
<option value="Home Furnishing Fabric">Home Furnishing Fabric</option>
</select>
Name: <input type="text"
 name="username" />
Time: <input type="text" name="time" />
<br>
<div id="weavediv">
</div>
</form>
 
</body>
</html>
 
 
-----------------------------------------------------------------------------------------
test.php
 
 
<?
include('db.php');
$fab = $_REQUEST['selfab'];
//echo $fab;
$sql=mysql_query("select * from weave where fabric='".$fab."'");
//echo "select * from fabric where fabric='".$fab."'";
?>
 
<select size="1" name="weavelist" name="weavelist" >
<?
while ($row = mysql_fetch_assoc($sql)) { ?>
<option value><?=$row['weave']?></option>
 <? } ?>
</select>
------------------------------------------------------------------
 
Keywords: Populating the drop down list using ajax.
Title
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6 Ajax results - make selectable and po…
 
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[+][-]07.05.2008 at 06:10AM PDT, ID: 21936894

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[+][-]07.05.2008 at 06:50AM PDT, ID: 21937081

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[+][-]07.05.2008 at 10:22AM PDT, ID: 21937702

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[+][-]07.07.2008 at 07:48AM PDT, ID: 21945514

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[+][-]07.07.2008 at 07:51AM PDT, ID: 21945533

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[+][-]07.07.2008 at 08:34AM PDT, ID: 21945963

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[+][-]07.07.2008 at 08:43AM PDT, ID: 21946038

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About this solution

Zones: Asynchronous Javascript and XML (AJAX), PHP Scripting Language
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Solution Provided By: hielo
Participating Experts: 3
Solution Grade: A
 
 
[+][-]07.07.2008 at 09:46PM PDT, ID: 21950971

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