ก๊องทำได้ 2 อันอะ แต่จะทำ List/Menu 3- 4 อัน แต่ละอันให้มันคงข้อมูลที่เราเลือกไว้ อะ
code
<body>
<form id="form1" name="form1" method="post" action="">
แบบรถ ::
<select name="price" id="price" onChange="location.href='menu_price.php?model_id='+this.value;" style="width:200px">
<option value=""> -- choose -- </option>
<?
include("conn/connection.php");
$sql ="SELECT DISTINCT model_car FROM model";
$dbquery=mysql_db_query($dbase,$sql);
//$result=mysql_fetch_array($dbquery);
//echo"$result[model_car]";
while($result=mysql_fetch_array($dbquery))
{
if($model_id == $result[model_car])
{
?>
<option VALUE="<?= $result[model_car]; ?>" selected="selected"><?=$result[model_car];?></option>
<? }else{?>
<option value="<?=$result[model_car]; ?>"><?=$result[model_car];?></option>
<?
}}
//-----------------------------------------------------------------
?>
</select>
<br/>
<br/>
รุ่น ::
<select name="option_car" id="option_car" tyle="width:200px">
<option value="">-- option car --</option>
<?
if($model_id !=""){
include("conn/connection.php");
$sql_op="SELECT * FROM model WHERE model_car='".$model_id."' ";
$dbquery_op=mysql_db_query($dbase,$sql_op);
$result_op_=mysql_fetch_array($dbquery_op);
echo"$result_op_[id_num]";
while($result_op=mysql_fetch_array($dbquery_op))
{
?>
<option value="<?=$result_op[option_car];?>"><?=$result_op[option_car]; $model_price=$result_op[option_car];?></option>
<?
} }
//-------------------------------------------------------------
?>
</select>
</form>
</body>