如何在不使用php提交数据的情况下传递下拉列表中的选定值？

Question

I'm using codeigniter .I have controller call feature.php and view call feature.php in view php i have drop-down list. I need to save my packages id in session when the user selected the drop-down.how can i do this without refreshing and submit data in my from ?

this is my HTML code in view php page

<div class="form-group">
<label for="lastname" class="control-label">Your Packages</label>
<?php if(isset($tourbuddy_packages)){?>
<select id="itemType_id" class="form-control input-sm" name="tripbuddy_PackageTitle" onChange="disp_text()">
    <?php foreach ($tourbuddy_packages as $packages) {?>
    <option value="<?php echo $packages['PackageID'] ?>">
        <?php echo $packages[ 'PackageTitle']?>
    </option>
    <?php } ?>
</select>
<input type="hidden" name="PackageID" id="country_hidden">
<?php } else { ?>
<select class="form-control input-sm" name="tripbuddy_PackageTitle">
    <option>Add Packages</option>
</select>
<?php } ?>
</div>

need quick help thanx

Answer 1

You can create a php file called save_package_to_session.php ;

<?php
if (!isset($_SESSION)) {
    session_start();    
}

$package_id = $_POST["id"];
$_SESSION["pkg_id"] = $package_id;
echo $_SESSION["pkg_id"];
?>

and js,

$("#itemType_id").change(function() {
    $.ajax({
        url : "save_package_to_session.php",
        method: "POST",
        data: "id=" + $(this).val(),
        success: function(response) {
            // handle
        }
    })  
});

Answer 2

I don't have more idea about php code but you make ajax request

 $("#itemType_id").change(function() {
   $.ajax({
     url : controller_action_url,
     method: "POST" /* use get or post*/,
     data: "package_id=" + $(this).val(),
     success: function(response) {
     }
  })  
});

simply, now you can get package_id params on controller and you can store in session.