Question

我想要一个php函数，当用52调用它时返回55。

我尝试了round()功能：

echo round(94, -1); // 90

它返回 90 ，但我想 95 。

感谢。

Answer 1

这可以通过多种方式实现，具体取决于您的首选舍入约定：

1。舍入到 next 5的倍数，排除当前数字

行为：50输出55,52输出55

function roundUpToAny($n,$x=5) {
    return round(($n+$x/2)/$x)*$x;
}

2。舍入到最近 5的倍数，包括当前数字

行为：50输出50,52输出55,50.25输出50

function roundUpToAny($n,$x=5) {
    return (round($n)%$x === 0) ? round($n) : round(($n+$x/2)/$x)*$x;
}

3。向上舍入到整数，然后到最接近的5

的倍数

行为：50输出50,52输出55,50.25输出55

function roundUpToAny($n,$x=5) {
    return (ceil($n)%$x === 0) ? ceil($n) : round(($n+$x/2)/$x)*$x;
}

Answer 2

除以5
round()（或ceil() ，如果您想要总结）
乘以5。

值5（分辨率/粒度）可以是任何值 - 在步骤1和3中都替换它

Answer 3

向下舍入：

$x = floor($x/5) * 5;

总结：

$x = ceil($x/5) * 5;

舍入到最接近（向上或向下）：

$x = round($x/5) * 5;

Answer 4

尝试我写的这个小功能。

function ceilFive($number) {
    $div = floor($number / 5);
    $mod = $number % 5;

    If ($mod > 0) $add = 5;
    Else $add = 0;

    return $div * 5 + $add;
}

echo ceilFive(52);

Answer 5

   echo $value - ($value % 5);

我知道这是一个老问题，但恕我直言使用模数运算符是最好的方法，并且比接受的答案更优雅。

Answer 6

来自Gears库

MathType::roundStep(50, 5); // 50
MathType::roundStep(52, 5); // 50
MathType::roundStep(53, 5); // 55

MathType::floorStep(50, 5); // 50
MathType::floorStep(52, 5); // 50
MathType::floorStep(53, 5); // 50

MathType::ceilStep(50, 5); // 50
MathType::ceilStep(52, 5); // 55
MathType::ceilStep(53, 5); // 55

来源：

public static function roundStep($value, int $step = 1)
{
    return round($value / $step) * $step;
}

public static function floorStep($value, int $step = 1)
{
    return floor($value / $step) * $step;
}

public static function ceilStep($value, int $step = 1)
{
    return ceil($value / $step) * $step;
}

Answer 7

乘以2，舍入为-1，除以2。

Answer 8

这是我的Musthafa's函数版本。这个更复杂，但它支持Float数字以及整数。要舍入的数字也可以是字符串。

/**
 * @desc This function will round up a number to the nearest rounding number specified.
 * @param $n (Integer || Float) Required -> The original number. Ex. $n = 5.7;
 * @param $x (Integer) Optional -> The nearest number to round up to. The default value is 5. Ex. $x = 3;
 * @return (Integer) The original number rounded up to the nearest rounding number.
 */
function rounduptoany ($n, $x = 5) {

  //If the original number is an integer and is a multiple of 
  //the "nearest rounding number", return it without change.
  if ((intval($n) == $n) && (!is_float(intval($n) / $x))) {

    return intval($n);
  }
  //If the original number is a float or if this integer is 
  //not a multiple of the "nearest rounding number", do the 
  //rounding up.
  else {

    return round(($n + $x / 2) / $x) * $x;
  }
}

我尝试了Knight，Musthafa中的功能，甚至是Praesagus的建议。他们不支持Float数字和来自Musthafa's＆amp;的解决方案。 Praesagus在某些数字中无法正常工作。请尝试以下测试编号并自行进行比较：

$x= 5;

$n= 200;       // D = 200     K = 200     M = 200     P = 205
$n= 205;       // D = 205     K = 205     M = 205     P = 210
$n= 200.50;    // D = 205     K = 200     M = 200.5   P = 205.5
$n= '210.50';  // D = 215     K = 210     M = 210.5   P = 215.5
$n= 201;       // D = 205     K = 205     M = 200     P = 205
$n= 202;       // D = 205     K = 205     M = 200     P = 205
$n= 203;       // D = 205     K = 205     M = 205     P = 205

** D = DrupalFever K = Knight M = Musthafa P = Praesagus

Answer 9

我这样做：

private function roundUpToAny(int $n, $x = 9)
{
    return (floor($n / 10) * 10) + $x;
}

试验：

assert($this->roundUpToAny(0, 9) == 9);
assert($this->roundUpToAny(1, 9) == 9);
assert($this->roundUpToAny(2, 9) == 9);
assert($this->roundUpToAny(3, 9) == 9);
assert($this->roundUpToAny(4, 9) == 9);
assert($this->roundUpToAny(5, 9) == 9);
assert($this->roundUpToAny(6, 9) == 9);
assert($this->roundUpToAny(7, 9) == 9);
assert($this->roundUpToAny(8, 9) == 9);
assert($this->roundUpToAny(9, 9) == 9);
assert($this->roundUpToAny(10, 9) == 19);
assert($this->roundUpToAny(11, 9) == 19);
assert($this->roundUpToAny(12, 9) == 19);
assert($this->roundUpToAny(13, 9) == 19);
assert($this->roundUpToAny(14, 9) == 19);
assert($this->roundUpToAny(15, 9) == 19);
assert($this->roundUpToAny(16, 9) == 19);
assert($this->roundUpToAny(17, 9) == 19);
assert($this->roundUpToAny(18, 9) == 19);
assert($this->roundUpToAny(19, 9) == 19);

Answer 10

function round_up($n, $x = 5) {
  $rem = $n % $x;
  if ($rem < 3)
     return $n - $rem;
  else
     return $n - $rem + $x;
}

Answer 11

我刚刚在20分钟内编写了这个功能，基于我在这里和那里找到的许多结果，我不知道它为什么会起作用或者它是如何工作的！：d

我主要想将货币数量从这个151431.1 LBP转换为150000.0 LBP。 （151431.1 LBP == ~100 USD）到目前为止工作得很好，但我试图让它与其他货币和数字兼容，但不确定它是否正常工作!!

/**
 * Example:
 * Input = 151431.1 >> return = 150000.0
 * Input = 17204.13 >> return = 17000.0
 * Input = 2358.533 >> return = 2350.0
 * Input = 129.2421 >> return = 125.0
 * Input = 12.16434 >> return = 10.0
 *
 * @param     $value
 * @param int $modBase
 *
 * @return  float
 */
private function currenciesBeautifier($value, int $modBase = 5)
{
    // round the value to the nearest
    $roundedValue = round($value);

    // count the number of digits before the dot
    $count = strlen((int)str_replace('.', '', $roundedValue));

    // remove 3 to get how many zeros to add the mod base
    $numberOfZeros = $count - 3;

    // add the zeros to the mod base
    $mod = str_pad($modBase, $numberOfZeros + 1, '0', STR_PAD_RIGHT);

    // do the magic
    return $roundedValue - ($roundedValue % $mod);
}

如果出现任何问题，请随意修改并修复

Answer 12

也许您也可以考虑使用这种衬板。更快！适用于$num >= 0和$factor > 0。

$num = 52;
$factor = 55;
$roundedNum = $num + $factor - 1 - ($num + $factor - 1) % $factor;

在PHP中舍入到最接近的五

12 个答案:

1。舍入到 next 5的倍数，排除当前数字

2。舍入到最近 5的倍数，包括当前数字

3。向上舍入到整数，然后到最接近的5