正如评论中指出的那样，这是一个范围界定问题.您的$d变量(mysqli实例)不在username_from_id的范围内.这是解决方法...

As pointed out in comments, this is a scoping issue. Your $d variable (mysqli instance) is not in scope within username_from_id. Here's how to fix it...

function username_from_id(mysqli $d, $id) {
    if (!$stmt = $d->prepare('SELECT username FROM users WHERE id = ? LIMIT 1')) {
        throw new Exception($d->error, $d->errno);
    }
    $stmt->bind_param('i', $id);

    if (!$stmt->execute()) {
        throw new Exception($stmt->error, $stmt->errno);
    }

    $stmt->bind_result($username);

    if ($stmt->fetch()) {
        return $username;
    }
    return null;
}

并这样称呼它

include 'file_where_function_is.php';
$id = 1;
echo username_from_id($d, $id); // assuming $d exists in this scope