php舍入值，尾随2位小数，5位的倍数

Question

I tried this function

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

echo roundUpToAny(4.52); // This show 5 instead of 4.55

echo roundUpToAny(5.1); // This show 5 instead of 5.10

How can i created such function

Answer 1

I think the following function does what you want:

function roundUpToAny($n, $x=5, $sf=2) {
  $scale = pow(10,$sf);
  return number_format(round(ceil($n*$scale / $x), $sf) * $x / $scale, $sf);
}

The first argument is the number to be converted, the second is the "rounding factor" (multiples of $x - you asked for 5 in the title of the question), the third is the number of figures after the decimal point.

Test:

echo roundUpToAny(1.23, 5, 2) ."\n";
echo roundUpToAny(4.54, 5, 2) ."\n";
echo roundUpToAny(5.101, 5, 3) ."\n";
echo roundUpToAny(1.19, 5, 3) ."\n";

Result:

1.25
4.55
5.105
1.190

UPDATE

You pointed out in a follow-up that the above code fails for inputs of 1.1 and 2.2 . The reason for this is the fact that these numbers are not exactly representable in double precision (strange though this may sound) . The following code compensates for this, by creating an intermediate value that is rounded to be an integer (and therefore can be represented exactly):

function roundUpToAny($n, $x=5, $sf=2) {
  $scale = pow(10,$sf);
  $temp = round($n * $scale, $sf); // get rid of small rounding errors
  return number_format(round(ceil($temp / $x), $sf) * $x / $scale, $sf);
}

Testing:

echo roundUpToAny(1.1, 5, 2) ."\n";

Result:

1.10

As expected.