四舍五入到最接近的尾数

我具有以下功能，将数字四舍五入到以$ nearest数字结尾的最接近的数字，我想知道是否有一种更优雅的方式做同样的事情.

I have the following function that rounds a number to the nearest number ending with the digits of $nearest, and I was wondering if there is a more elegant way of doing the same.

/**
 * Rounds the number to the nearest digit(s).
 *
 * @param int $number
 * @param int $nearest
 * @return int
 */

function roundNearest($number, $nearest, $type = null)
{
    $result = abs(intval($number));
    $nearest = abs(intval($nearest));

    if ($result <= $nearest)
    {
        $result = $nearest;
    }

    else
    {
        $ceil = $nearest - substr($result, strlen($result) - strlen($nearest));
        $floor = $nearest - substr($result, strlen($result) - strlen($nearest)) - pow(10, strlen($nearest));

        switch ($type)
        {
            case 'ceil':
                $result += $ceil;
            break;

            case 'floor':
                $result += $floor;
            break;

            default:
                $result += (abs($ceil) <= abs($floor)) ? $ceil : $floor;
            break;
        }
    }

    if ($number < 0)
    {
        $result *= -1;
    }

    return $result;
}

一些例子:

roundNearest(86, 9); // 89
roundNearest(97, 9); // 99
roundNearest(97, 9, 'floor'); // 89

提前谢谢！

PS:这个问题不是舍入到最近倍数的.