Populate Dropdown list depending on another Dropdown selection using Ajax

Question

I need to select an id depending on another id. I made by myself something but it is not working, so I am posting the CODE, maybe i have something wrong in it. Select the first option a vehicle, and in the second option select a model of the vehicle that I selected. Thanks!

//The config file

<?php

$host = 'localhost';         
$utilizator = 'stud';         
$parola = 'stud';         
$numebd = 'autodealer';      

$conn = mysql_connect($host, $utilizator, $parola);
if (!$conn) { echo '<h4>Connected!</h4>'; }

if (!mysql_select_db($numebd, $conn)) { 
   echo '<h4>Couldnt connect database : '. mysql_errno(). ' : '. mysql_error().'</h4>'; 
}

mysql_set_charset('utf8', $conn);
?>


//The HTML code

<h2><strong>Vehicle</strong> data</h2>
    <div class="select_wrapper">
    <form action="" method="post" name="anunt" id="f_anunt">
        <label><span>* </span><strong>Manufacturer:</strong></label>
        <select class="select_5" name="marca" id="marca">                           
            <?php
                $result = mysql_query("SELECT id_marca, denumire FROM marci ORDER BY ordine");
                while($row = mysql_fetch_array($result)) {
                    echo ("<option value='".$row['id_marca']."'>".$row['denumire']."</option>");
                }
            ?>
        </select>
        </div>
        <div class="select_wrapper">
        <label><span>* </span><strong>Model: </strong></label>
            <select class="select_5" name="model" id="model">
                <?php                                   
                    $result = mysql_query("SELECT id_model, denumire FROM modele WHERE id_marca=1");
                    while ($row = mysql_fetch_array($result)) {
                        echo ("<option value='".$row['id_model']."'>".$row['denumire']."</option>");
                    }
                ?>
            </select>
        </div>

//The PHP file

<?php
include_once("config.php");

$id_marca = $_POST['id_marca'];

$sir_sql="SELECT id_model, denumire FROM modele WHERE id_marca=$id_marca ";

if ($conn())
    {               
        $result=mysql_query($sir_sql,$conn);

        while ($row = mysql_fetch_array($result))       
           echo ("<option value='".$row['id_model']."'>".$row['denumire']."</option>");

    }
    else 
        die('Comanda incorecta:<br> ' . mysql_error()); 
?>

//The Javascript code

$("#marca").change(get_model);

    function get_model(e)
{
    $.post("ajax_get_model.php",{id_marca:$(this).val()},function(data,status)
        {   
            $("#model").html(data); // pun modelele 
        });
}

Beware, as it stands this contains a possible SQL Injection attack. — Ryan Leach
– Ryan Leach, Commented Feb 3, 2014 at 9:31

Community · Accepted Answer · 2017-05-23 10:27:30Z

1

There are several post on SO about this topic. Please, read these topics first, hope these will help-

Finally, you can check the tutorial here to populate dropdown based on dropdown selection...

Tutorial Here | Demo Here

edited May 23, 2017 at 10:27

CommunityBot

11 silver badge

answered Feb 3, 2014 at 9:41

user2160110

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3 Comments

MariusB Over a year ago

Thank you! That was very helpful. The problem was in the PHP file.. if ($conn) instead of if ($conn()) .

MariusB Over a year ago

Yeah, sure. And I like your links. Made me understand much more about ajax.

MariusB Over a year ago

As soon as I get 15 reputation, that is the condition I guess!

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