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how to load values from form to jquery

Posted on 2009-04-04
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Last Modified: 2012-06-27
am having 2 a select box with 2 options....

when i i select an option1 and click submit, i get values relating to the first option in the grid. But when i change to option2  the values does not change.

i need to refresh again and when i choose option2 this time i get the corresponding values and again when i try to change to option1 same prob as above.....pls help
<?
mysql_connect("localhost","root","");
mysql_select_db("lab")or die("CANNOT CONNECT To DATABASE".mysql_error());
?>
 
<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd">
<html xmlns="http://www.w3.org/1999/xhtml">
<head>
	<meta http-equiv="content-type" content="text/html; charset=iso-8859-1" />
<title>Flexigrid</title>
<link rel="stylesheet" type="text/css" href="css/flexigrid.css" />
<script type="text/javascript" src="jquery-1.2.3.pack.js"></script>
<script type="text/javascript" src="flexigrid.js"></script>
 
<script type="text/javascript">
 
function getXMLHTTP() { //fuction to return the xml http object
		var xmlhttp=false;	
		try{
			xmlhttp=new XMLHttpRequest();
		}
		catch(e)	{		
			try{			
				xmlhttp= new ActiveXObject("Microsoft.XMLHTTP");
			}
			catch(e){
				try{
				xmlhttp = new ActiveXObject("Msxml2.XMLHTTP");
				}
				catch(e1){
					xmlhttp=false;
				}
			}
		}
		 	
		return xmlhttp;
    } 
 
 
$(document).ready(function(){
 
$('form').submit(function() {
 
  var fname=$('#test_n').val();
	
	$("#flex1").flexigrid
			(
			{
			
			url: 'post2.php?type='+fname,
			dataType: 'json',
			colModel : [
			
				{display: 'Test ID', name : 'test_id', width : 40, sortable : true, align: 'center'},
				{display: 'Test Code', name : 'tst_code', width : 80, sortable : true, align: 'left'},
				{display: 'Test Name', name : 'test_name', width : 180, sortable : true, align: 'left'},
				{display: 'Test Dept ID', name : 'tst_dep_id', width : 70, sortable : true, align: 'center'},
				{display: 'Test Sample ID', name : 'tst_sample_id', width : 70, sortable : true, align: 'center' },//hide: true},
				{display: 'Normal Value', name : 'normal_val', width : 80, sortable : true, align: 'left',hide:true}
				//{display: 'Number', name : 'num', width : 80, sortable : true, align: 'right'}
				],
			
			searchitems : [
				{display: 'Test Code', name : 'tst_code'},
				{display: 'Test Name', name : 'test_name', isdefault: true}
				],
			sortname: "test_id",
			sortorder: "asc",
			usepager: true,
			title: 'TEST LIST	',
			useRp: true,
			rp: 10,
			showTableToggleBtn: true,
			width: 700,
			height: 255
			}
			);   
	
return false;
});
});
 
</script>
</head>
<br /><br />
 
<body align="center" bgcolor="#ffffff">
<form method='post' action='<? $_SERVER['PHP_SELF'] ?>' onsubmit='return validate_form();'>
 
<div id="statediv">Test Name:&nbsp;&nbsp;&nbsp;<select id="test_n" name="test_name"  >
	<option value="test" >All</option>
    <option value="test1" >All2</option>
    
        </select>
		<input type="submit" value="submit"></div>
<?
//echo $_REQUEST['sortname'];
echo "<table id='flex1' style='display:none' ></table>"
?>
 
</form>
<br /><br />
 
</body>
</html>

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Question by:whspider
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3 Comments
 
LVL 5

Expert Comment

by:prokvk
ID: 24067589
How about posting your validate_form() function ?
0
 

Author Comment

by:whspider
ID: 24067664
am not using validation....

i just want the grid the have values based on the selected options
0
 

Accepted Solution

by:
whspider earned 0 total points
ID: 24144922
thanks i found it myself.i included the following and it worked

var fname=$('#test_n option:selected').val();
var fname1=
jQuery('#flex1').flexOptions({newp:1, params:[{name:'type', value: fname},{name:'sal',value:fname1}]});
jQuery("#flex1").flexReload();
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