Предупреждение: mysqli_stmt_close () ожидает, что параметр 1 будет mysqli_stmt, логическое значение

Question

Предупреждение: mysqli_stmt_close () ожидает, что параметр 1 будет mysqli_stmt, с логическим значением

Я получил эту ошибку, не знаю почему, потратил 3 часа на проверку всего и не могу найти проблему

// Prepare a select statement
$sql = "SELECT countries.name AS country_name, customers.id, customers.name AS customer_name, surname, phone, email, address1, code FROM customers WHERE id = ? JOIN countries ON country_code = code";

if($stmt = mysqli_prepare($link, $sql)){
    // Bind variables to the prepared statement as parameters
    mysqli_stmt_bind_param($stmt, "i", $param_id);

    // Set parameters
    $param_id = trim($_GET["id"]);

    // Attempt to execute the prepared statement
    if(mysqli_stmt_execute($stmt)){
        $result = mysqli_stmt_get_result($stmt);

        if(mysqli_num_rows($result) == 1){
            /* Fetch result row as an associative array. Since the result set
            contains only one row, we don't need to use while loop */
            $row = mysqli_fetch_array($result, MYSQLI_ASSOC);

            // Retrieve individual field value
            $name = $row["customer_name"];
            $address1 = $row["address1"];
            $salary = $row["code"];
        } else{
            // URL doesn't contain valid id parameter. Redirect to error page
            header("location: error.php");
            exit();
        }

    } else{
        echo "Oops! Something went wrong. Please try again later.";
    }
}

// Close statement
var_dump($stmt);
mysqli_stmt_close($stmt);

// Close connection
mysqli_close($link);}