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PHP 中四舍五入到最接近的五的倍数

发布于 2024-10-01 00:25:26 字数 188 浏览 4 评论 0原文

我想要一个 php 函数，当用 52 调用它时返回 55。

我尝试了 round() 函数：

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

它返回 90 但我想要 95。

谢谢。

原文

I want a php function which returns 55 when calling it with 52.

I've tried the round() function:

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

It returns 90 but I want 95.

Thanks.

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评论（12）

天涯沦落人 2024-10-08 00:25:26

这可以通过多种方式完成，具体取决于您首选的舍入约定：

1. 舍入到下一个 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. 舍入到最接近 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. 向上舍入到一个整数，然后是最接近的 5 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;
}

This can be accomplished in a number of ways, depending on your preferred rounding convention:

1. Round to the next multiple of 5, exclude the current number

Behaviour: 50 outputs 55, 52 outputs 55

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

2. Round to the nearest multiple of 5, include the current number

Behaviour: 50 outputs 50, 52 outputs 55, 50.25 outputs 50

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

3. Round up to an integer, then to the nearest multiple of 5

Behaviour: 50 outputs 50, 52 outputs 55, 50.25 outputs 55

function roundUpToAny($n,$x=5) {
    return (ceil($n)%$x === 0) ? ceil($n) : round(($n+$x/2)/$x)*$x;
}
回复 原文
宫墨修音 2024-10-08 00:25:26
  1. 除以 5
  2. round() （或 ceil() 如果您想始终向上舍入）
  3. 乘以 5。

值 5（分辨率/粒度）可以是任何值 - 在步骤 1 和步骤 3 中替换它

所以总而言之：

    $rounded_number = ceil( $initial_number / 5 ) * 5
  1. Divide by 5
  2. round() (or ceil() if you want to round up always)
  3. Multiply by 5.

The value 5 (the resolution / granularity) can be anything — replaced it in both step 1 and 3

So in summary:

    $rounded_number = ceil( $initial_number / 5 ) * 5
回复 原文
瞄了个咪的 2024-10-08 00:25:26

向下舍入：

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

向上舍入：

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

四舍五入到最接近的位置（向上或向下）：

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

Round down:

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

Round up:

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

Round to closest (up or down):

$x = round($x/5) * 5;
回复 原文
幼儿园老大 2024-10-08 00:25:26
   echo $value - ($value % 5);

我知道这是一个老问题，但恕我直言，使用模运算符是最好的方法，并且比公认的答案优雅得多。

   echo $value - ($value % 5);

I know it's an old question, but IMHO using modulus operator is the best way, and far more elegant than the accepted answer.

回复 原文
拥抱影子 2024-10-08 00:25:26

试试我写的这个小函数。

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);

Try this little function I wrote.

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);
回复 原文
冷弦 2024-10-08 00:25:26

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;
}

From Gears library

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

Source:

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;
}
回复 原文
聊慰 2024-10-08 00:25:26

乘以 2，舍入为 -1，除以 2。

Multiply by 2, round to -1, divide by 2.

回复 原文
甜中书 2024-10-08 00:25:26

这是我的 Musthafa 函数版本。这个比较复杂，但它支持浮点数以及整数。要舍入的数字也可以位于字符串中。

/**
 * @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 的功能，甚至是 Musthafa 的建议="/users/58013/praesagus">Praesagus。他们不支持浮点数字以及 Musthafa's & 的解决方案。 Praesagus 在某些数字中无法正常工作。尝试以下测试数字并自己进行比较：

$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

Here is my version of Musthafa's function. This one is more complex but it has support for Float numbers as well as Integers. The number to be rounded can also be in a string.

/**
 * @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;
  }
}

I tried the functions from Knight, Musthafa and even the suggestion from Praesagus. They don't have support for Float numbers and the solutions from Musthafa's & Praesagus do not work correctly in some numbers. Try the following test numbers and do the comparison yourself:

$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
回复 原文
赤濁 2024-10-08 00:25:26

我这样做：

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

测试：

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);

I do it like this:

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

Tests:

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);
回复 原文
再见回来 2024-10-08 00:25:26
function round_up($n, $x = 5) {
  $rem = $n % $x;
  if ($rem < 3)
     return $n - $rem;
  else
     return $n - $rem + $x;
}

function round_up($n, $x = 5) {
  $rem = $n % $x;
  if ($rem < 3)
     return $n - $rem;
  else
     return $n - $rem + $x;
}
回复 原文
最好是你 2024-10-08 00:25:26

我刚刚在 20 分钟内写了这个函数，基于我到处发现的许多结果，我不知道它为什么起作用或者它是如何工作的！ :D

我主要感兴趣的是将货币数字从 151431.1 LBP 转换为 150000.0 LBP。 (151431.1 LBP == ~100 USD) 到目前为止，它工作得很好，但是我试图让它以某种方式与其他货币和数字兼容，但不确定它是否工作正常！

/**
 * 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);
}

如果有什么问题欢迎修改和修复

I just wrote this function in 20 min, based on many results I found here and there, I don't know why it works or how it works!! :D

I was mainly interested in converting currency numbers from this 151431.1 LBP to 150000.0 LBP. (151431.1 LBP == ~100 USD) which works perfectly so far, however I tried to make it somehow compatible with other currencies and numbers, but not sure if it works fine!!

/**
 * 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);
}

Feel free to modify it and fix it if there's anything wrong

回复 原文
猫烠⑼条掵仅有一顆心 2024-10-08 00:25:26

也许您也可以考虑这款衬里。更快了！适用于 $num >= 0 和 $factor > 0 。

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

Probably you can also consider this one liner. It's faster! Works for $num >= 0 and $factor > 0.

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