Question

我有以下功能将数字四舍五入到最接近的数字，以$ nearest 结尾，我想知道是否有更优雅的方式做同样的事。

/**
 * 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

提前致谢！

PS：这个问题不关于舍入到最近的多个。

Answer 1

这对我有用：

function roundToDigits($num, $suffix, $type = 'round') {
    $pow = pow(10, floor(log($suffix, 10) + 1));
    return $type(($num - $suffix) / $pow) * $pow + $suffix; 
};

$type应为“ceil”，“floor”或“round”

Answer 2

我认为这应该有效，而且对我来说更优雅，至少：

function roundNearest($number, $nearest, $type = null)
{
  if($number < 0)
    return -roundNearest(-$number, $nearest, $type);

  $nearest = abs($nearest);
  if($number < $nearest)
    return $nearest;

  $len = strlen($nearest);
  $pow = pow(10, $len);
  $diff = $pow - $nearest;

  if($type == 'ciel')
    $adj = 0.5;
  else if($type == 'floor')
    $adj = -0.5;
  else
    $adj = 0;

  return round(($number + $diff)/$pow + $adj)*$pow - $diff;
}

修改：从负面输入中添加了我想要的想法。

四舍五入到最近的结尾数字

2 个答案: