ต้องการแสดงข้อมูลที่ดึงจากฐานข้อมูลมาโชว์ ที่ listbox เมื่อเลือกข้อมูลใน listbox จากนั้นให้แสดงข้อมมูลที่ดึงจากฐานข้อมูลอีกอันหนึ่ง

<select name="test">
   <option value="">เลือก</option>
   <?php
      $sql  = "SELECT * FROM tbtest";
      $result = mysql_query($sql);
      while($show = mysql_fetch_array($result)){
   ?>
      <option value="<?=$show['test_id']?>"><?=$show['test_name']?></option>
   <?php } ?>
</select>

<meta http-equiv="Content-Type" content="text/html; charset=utf-8" />
<script src="http://code.jquery.com/jquery-latest.js"></script>
<script type="text/javascript">
	$(document).ready(function() {
		$('#rawmaterial').change(function(){
			var materail = $(this);
			var slices = $('option:selected',materail).attr('rel').split(",");
			material.parents('tr').find('input.cp').val(slices[0]);
			material.parents('tr').find('input.kcalkg').val(slices[1]);
			material.parents('tr').find('input.cal').val(slices[2]);			
		});
	});
</script>
<table><tr><td>
<select name='rawmaterial' id='rawmaterial'>
<?php
	$sql = "select * from material";
	$res = mysql_query($sql) or die(mysql_error());
	while($data = mysql_fetc_assoc($res)){
		echo "<option value='{$data["id"]}' rel='{$data["cp"]},{$data["kcalkg"]},{$data["cal"]}'>{$data["material"]}</option>\n";
	}
?>
</select>
</td>
<td>
<input type='text' name='cp[]' class='cp' />
</td>
<td>
<input type='text' name='kcalkg[]' class='kcalkg' />
</td>
<td>
<input type='text' name='cal[]' class='cal' />
</td>
</tr>
</table>

<script src="jquery.10.js"></script>
<script>
	$(function(){
		$('#sid').change(function(){
			var val=$(this).val();
			$.ajax({url:"check.php",
				async:false,
				cache:false,
				global:false,
				type:'POST',
				data:'id='+val,
				success: function(result){
					var obj=jQuery.parseJSON(result);
					$.each(obj,function(key,val){
						$('#demand_nut').val(val['demand_nut']);	
					});
				}
			});	
		});	
	});
</script>
<form name="form" id="form">
	<select id="sid">
    <option value="">Select</option>
	<?PHP
        $select=mysql_query("select * from weight_range") or die (mysql_error());
		while($result=mysql_fetch_array($select)){
	?>
    <option value="<?=$result['WR_id'];?>"><?=$result['WR'];?></option>
    <?PHP } ?>
	</select>
    <input type="number" id="demand_nut" />
</form>

<?PHP
	$id=mysql_real_escape_string($_POST['id']);
	$arrResult=array();
	$select=mysql_query("select * from divide WHERE WR_id='".$id."'") or die (mysql_error());
	$numFields=mysql_num_fields($select);
	while($result=mysql_fetch_array($select)){
		$arrCol=array();
		for($i=0;$i<$numFields;$i++){
			$arrCol[mysql_field_name($select,$i)]=$result[$i];	
		}
		array_push($arrResult,$arrCol);
	}
	
	echo json_encode($arrResult);
?>