Inserted data are not going to my database

Question

I am trying to upload a image in my website with some data. The image that I am trying to upload is being uploaded in my image folder but the data is not going in by database .And it's showing an error in mysqli_stmt_bind_param . I couldn't figure out an error. can someone please help me ?

if (isset($_POST['submit1'])) {
    $cakename = $_POST['cakeName'];

    $Beschreibung = $_POST['beschreibung'];
    $Preis = $_POST['preis'];

    $file = $_FILES['image'];

    $fileName = $file["name"];
    $filetype = $file["type"];
    $fileTempName = $file["tmp_name"];
    $fileError = $file["error"];
    $fileSize = $file["size"];

    $fileExt = explode(".", $fileName);
    $fileActualExt = strtolower(end($fileExt));

    $allowed = array("jpg", "jpeg", "png");

    if (in_array($fileActualExt, $allowed)) {
        if ($fileError === 0) {
            if ($fileSize < 2000000) {
                $imageFullName = $cakename.".". uniqid("", true).".". $fileActualExt;
                $fileDestination = "../Bake_my_cake1/image/".$imageFullName;

                include_once 'lib/db_connector.php';
                $db = dbconnect();

                if (empty($Beschreibung) || empty($Preis)) {
                    header("Location: ../Bake_my_cake1/Profile_baker1.php?upload=empty");
                    exit();
                } else {
                    $sql = "SELECT * FROM Cake;";
                    $stmt = mysqli_stmt_init($db);
                    if (!mysqli_stmt_prepare($stmt, $sql)) {
                        echo "SQL statement failed 1";
                    } else {
                        mysqli_stmt_execute($stmt);
                        $result = mysqli_stmt_get_result($stmt);
                        $rowCount = mysqli_num_rows($result);
                        $setImageOrder = $rowCount + 1;

                        $sql = "INSERT INTO Cake (CakeName, Beschreibung, Preis, orderGallery) VALUES(?, ?, ?, ?);";

                        if (!mysqli_stmt_prepare($stmt, $sql)) {
                            echo "SQL statement failed 2";
                        } else {
                            mysqli_stmt_bind_param($stmt, "ssss", $imageFullName, $Beschreibung, $Preis, $setImageOrder);
                            mysqli_stmt_execute($stmt);
                            echo "error 1";
                            move_uploaded_file($fileTempName, $fileDestination);

                            header("Location ../Bake_my_cake1/Profile_baker1.php?upload=sucess");
                        }
                    }
                }
                dbclose($db);
            } else {
                echo"File Size is too Big";
                exit();
            }
        } else {
            echo"You have an error";
            exit();
        }
    } else {
        echo"You need to upload a proper file type";
        exit();
    }
}

?>

else{
    mysqli_stmt_bind_param($stmt,"ssss", $imageFullName, $Beschreibung, $Preis, $setImageOrder);
    mysqli_stmt_execute($stmt);
    echo "error 1";
    move_uploaded_file($fileTempName,$fileDestination);

    header("Location ../Bake_my_cake1/Profile_baker1.php?upload=sucess");

it is showing error 1 .

Answer 1

according to your variable name i'm assuming that your function dbconnect() returns database object. And you're passing that $db in mysqli_stmt_init instead of DB connection object.

ex.

If you are returning database connection object and storing it in $con then your code will be look like $stmt = mysqli_stmt_init($con) .

Using this code you will not see "SQL statement failed 1" message and your code will be executed as expected.