PHP function Not Working As Expected From functions.php Include File

Question

I have a function in a functions.php file which updates a download count in a MySQL database via php. I cannot fathom out why it isn't working.

Below is the code that invokes the function. I'm also not getting any errors showing in my error logs.

if($db_image_id) {
    downloadCounter($connection, $db_image_id);
}

Below is the function this is linked to in functions.php. Other functions from this file work so I know it is being included correctly. Note: the $connection variable is the connection from a db.php file

function downloadCounter($connection, $imageID) {
    if (isset($_POST['download'])) {

        // value from hidden form element
        $imageID = $_POST['image-id'];

        try {
            $sql = "UPDATE imageposts SET downloads = downloads +1 WHERE image_id = :image_id";
            $stmt = $connection->prepare($sql);

            $stmt->execute([
                    ':image_id' => $imageID
            ]);

        } catch (PDOException $e) {
            echo "Error: " . $e->getMessage();
        }
    }
}

When I do var_dump($db_image_id) when there are three images on the page I get the following when the page loads - it only references one of the 3 images by id (namely id of 164):

`string(3) "164"`

The PHP for the page is below.

<?php

include "functions.php";

if (isset($_GET['username'])) {
    $username = $_GET['username'] 
} else { 
    header("Location: login.php");
    exit;
}
?>

<aside class="masonry-grid">
    <?php

        $stmt = $connection->prepare("SELECT * FROM imageposts WHERE username = :username");
        $stmt->bindParam(':username', $username);
        $stmt->execute();

        while ($image_row = $stmt->fetch()) {

        $db_image_id = escape($image_row['image_id']);
        $db_image_title = escape($image_row['image_title']);
        $db_image_tags = escape($image_row['image_tags']);
        $db_image_filename = escape($image_row['filename']);
        $db_ext = escape($image_row['file_extension']);

    ?>
    <figure>
        <form method="post">

        <!-- HTML OUTPUT OF IMAGES THAT INCLUDES ABOVE PHP VARIABLES -->

        <!-- DOWNLOAD BUTTON THAT TRIGGERS THE DOWNLOAD COUNTER FUNCTION  -->
            <button type="submit" name="download">Download</button>
            <input type="hidden" name="image-id" value="<?php echo $db_image_id; ?>">
        </form>
    </figure>
    
<?php } ?>
            
<?php

    // theoretically updates a download count in database via the functions.php function but isn't working
    if(isset($db_image_id)) {
        downloadCounter($connection, $db_image_id);
    }
?>
</aside>

Answer 1

if(isset($db_image_id)) {
    downloadCounter($connection, $db_image_id);
}

There may be something wrong. The var $db_image_id is the last value of image_row. We could use $_POST['image-id']. For example:

if(!empty($_POST['image-id'])) {
    downloadCounter($connection, $_POST['image-id']);
}

btw, the function downloadCounter() could be improved too. You can see another answer of Onki Hara.

Answer 2

Maybe you should call the function like this:

if(!empty($_POST['download'])) {
    downloadCounter($connection);
}

And your downloadCounter function could look like this:

function downloadCounter($connection) {

    if (empty($_POST['image-id'])) {
        return;
    }

    // value from hidden form element
    $imageID = $_POST['image-id'];

    try {
        $sql = "UPDATE imageposts SET downloads = downloads +1 WHERE image_id = :image_id";
        $stmt = $connection->prepare($sql);

        $stmt->execute([
                ':image_id' => $imageID
        ]);

    } catch (PDOException $e) {
        echo "Error: " . $e->getMessage();
    }
}