Auto-fill input form with PHP, MySQL and jQuery

Question

I had a problem with my auto-fill input form, with PHP MySQL and jQuery of course. So I have the select option form and some 'disabled' textboxes. And I want when I select the one of the option, the textboxes is auto-fill with value from the database

So there is my whole code :

PHP (Database connection) - I put it above my <html> code

<?php
    $host = 'localhost';
    $user = 'root';
    $pass = '';
    $db = 'db_mydatabase';

    mysql_connect($host, $user, $pass);
    mysql_select_db($db);

    $user = '';
    $query = mysql_query("SELECT * FROM tb_payment") or die (mysql_error());
?>

HTML

<select name="student" class="input-data">
    <option value="">-- SELECT STUDENT --</option>
    <!-- Do loop data from database -->
    <?php while ($row = mysql_fetch_object($query)): ?>
        <option value="<?php echo $row->id_user ?>">
            <?php echo $row->name ?>
        </option>
    <?php endwhile; ?>
</select>
<br />
<table border=1>
    <tr>
        <th>ID</th>
        <td><input type="text" class="output-id" disabled /></td>
    </tr>
    <tr>
        <th>Name</th>
        <td><input type="text" class="output-name" disabled /></td>
    </tr>
    <tr>
        <th>Total Payment</th>
        <td><input type="text" class="output-total" disabled /></td>
    </tr>
</table>

jQuery :

$(document).ready(function(){
    $(".input-data").on("change", function(){
        <?php $id_user = "$(this).val()"; ?>

        <?php $data = mysql_query("SELECT * FROM tb_payment WHERE id_user = '$id_user'") or die (mysql_error()); ?>
        <?php while($row = mysql_fetch_object($data)): ?>
            $(".output-id").val(<?php $row->id_user ?>);
            $(".output-name").val(<?php $row->name ?>);
            $(".output-total").val(<?php $row->total ?>);
        <?php endwhile; ?>
    });
});

But when i try to select the option, the values are wouldn't appear. Can someone help me?

What if you remove this extra < in the 4th line of the jQuery part? — Phiter
– Phiter, Commented Feb 10, 2016 at 2:06
Do you have error reporting on? This should be causing issues while (row - mysql_fetch_object($query)):.. — chris85
– chris85, Commented Feb 10, 2016 at 2:06
@chris85 It is no error that reported but when I open my firebug, the java script code its just $(document).ready(function(){ $(".input-data").on("change", function(){}); }); — imbagila
– imbagila, Commented Feb 10, 2016 at 2:26

Ahmed Abu Taha · Accepted Answer · 2016-02-10 03:03:01Z

What about creating new page name process.php contain:

require_once "connectDB.php";

header('Content-type: application/json; charset=utf-8');

if(isset($_POST['one'])){
$json = array();
$id =  trim($_POST['one']);
$query = "SELECT id, name, total FROM tb_payment WHERE id_user = ?";
$statement = $databaseConnection->prepare($query);
$statement->bind_param('s', $id);
$statement->execute();
$statement->bind_result($nId, $nName, $nTotal);
while ($statement->fetch()){
   $user=array('id'=>$nId,'name'=>$nName,'total'=>$nTotal);
    array_push($json,$user);
}
echo json_encode($json, true);

 }

And the Jquery:

$(document).ready(function(){
    $(".input-data").on("change", function(){
            var id = $(".input-data").val();
            var data = 'one=' + id;
            $.ajax({
                type: "POST",
                url: "process.php",
                data: data,
                dataType: 'json',
                success: function (data) {
                    if (data) {
                        for (var i = 0; i < data.length; i++) { //for each user in the json response
                            $(".output-id").val(data[i].id);
                $(".output-name").val(data[i].name);
                $(".output-total").val(data[i].total);
                        } // for

                    } // if
                } // success
            }); // ajax
    });
});

Cavid · Accepted Answer · 2016-04-29 13:52:40Z

0

You have to print

 $(".output-id").val(<?php print $row->id_user ?>);
 $(".output-name").val(<?php print $row->name ?>);
 $(".output-total").val(<?php print $row->total ?>);

answered Apr 29, 2016 at 13:52

Cavid

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