使用Ajax的动态列表

Question

my website is comugg.com. Im trying to make a dynamic dropdown according to this portion of my website.

https://gyazo.com/bc68919205d61b2022543d1aef81ca36

Anyways i got the first and 2nd level to work. Im struggling on my third level. i've narrowed the problem down to these two areas within my code. If anyone can give me a second eye on this i would appreciate it.

        <div id="model">
    <label>Model</label>
        <select name="model" id="modellist" onchange="getId2(this.value);">
            <option value="">Select Model</option>
        </select>
    </div>

            <div id="year">
    <label>Year</label>
        <select name="year" id="yearlist" >
            <option value="">Select Year</option>
        </select>
    </div>


    <script>
        function getId(val){
                jQuery.ajax({
                method: "POST",
                url: "getdata.php",
                data: "make="+val,
                success:function(data){
                    jQuery("#modellist").html(data);
                }
                });


        }
        function getId2(val){
                jQuery.ajax({
                method: "POST",
                url: "getdata2.php",
                data: "model="+val,
                success:function(data){
                    jQuery("#yearlist").html(data);
                }
                });

        }

    </script>

and the next part of the getdata2.php is

<?php
include_once "connection.php";
$make = $_POST["year"];


if (!empty($_POST["year"])) {
    $make = $_POST["year"];
    echo $query = "SELECT Distinct Year FROM websitemasterlist where Model='".$model."'"; 
    $results = mysqli_query($conn, $query);
    ?>
    <option value="">Select Year</option>
    <?php
    foreach ($results as $info){
    ?>
    <option value="<?php echo $info["Year"]; ?>"><?php echo $info["Year"]; ?></option>
    <?php
    }
}
?>

any help would be appreciated thank you.

Answer 1

i do not see any year post data so update this code , hope it will work, in ajax code you use post data as

data: "model="+val,

so in your php code need to use same post value

if (!empty($_POST["model"])) {
    $make = $_POST["model"];
    echo $query = "SELECT Distinct Year FROM websitemasterlist where Model='".$make."'"; 
    $results = mysqli_query($conn, $query);
    ?>
    <option value="">Select Year</option>
    <?php
    foreach ($results as $info){
    ?>
    <option value="<?php echo $info["Year"]; ?>"><?php echo $info["Year"]; ?></option>
    <?php
    }
}
?>