Displaying from drop down menu into table

[gezzam]

Hi all, just new to php.

To get to the point, I populate a dropdown menu from a database in MySQL. What I need to be able to do is when the user selects an item from the dropdown menu I want to display the cost of the item in a table next to the table contaning the drop down menu.

The code is here :

<?
mysql_select_db("ecafe");
$result = mysql_query("select * from bev_hot");
$options="";

while($r=mysql_fetch_array($result))
{
	$item=$r["item"];
	$options.="<OPTION VALUE=\"$item\">".$item.'</option>';
}
?>

<tr>
  <td width=216>
  <select name=products>
  	<option value=0>Choose a Hot Drink
	<?=$options	?>
	</option>
  </select>
  </td>
</form>

As you can see the dropdown list is populated from the "bev_hot" table in MySQL. In this table there is also a field called cost which is the cost of the item (in this case a hot drink)

What I need to do is once the user selects the beverage I need to display the cost of the drink in the table directly next to it.

The table fields that I need to access are "item" (Product Description) and "cost" (Cost of product)

Does this make any sense at all and if so is it at all possible???

[TTT]

<script language="javascript" type="text/javascript">
function sPrice() {
    document.getElementById("showhere").innerHTML = document.yourFormName.products.options[document.yourFormName.products.selectedIndex].value;

}

</script>

<tr> 
  <td width=216> 
  <select name=products onchange="javascript : sPrice();"> 
      <option value=0>Choose a Hot Drink 
    <?=$options    ?> 
    </option> 
  </select> 
  </td> 
  <td><div id="showhere"></div>
  <td>
</form> 

[gezzam]

Almost there.......thanks for the help, much appreciated.

The only problem now is I don't get the cost field displayed in the table, it just repeats whatever field that I selected from the drop down menu.

At least I am alot further down the track than I was before....lol