PHP - How to validate float number has maximum 2 digits - php

Question

I have to validate if a float number has maximum two digits.
I've tried a lot of methods but all fails in more or lase cases.
Last of them were:
//fails for 2638655.99
private function hasMoreThanTwoDecimals(string $number): bool
{
$number = abs($number);
$intPart = floor($number);
$floatPart = $number - $intPart;
return (strlen($floatPart) > 4);
}
OR
//fails for 36.62
private function hasMoreThanTwoDecimals(string $number): bool
{
return $number * 100 - floor($number * 100) > 0.00001;
}
What other methods do you use?

Answer 1

You can't determine the exact number of decimals with the float datatype, because the internal representation is binary. In binary, fx. 0.1 can not be represented exactly. That's why loops always should have integer increments.
for ($i = -1; $i < 1; $i += 0.1) {
if ($i == 0) {
echo "Zero is here!";
}
}
will never say "Zero is here!" because of binary rounding issues.
Using an Epsilon
You already tried to use an epsilon (a very small value) for thesholding (here a refactored version of your function):
private function hasMoreThanTwoDecimals(string $number): bool
{
$epsilon = 0.00001;
return fmod($number * 100, 1.0) > $epsilon;
}
but fails for some values. In that case, you need to increase your epsilon value.
String Arithmetic
The more precise way is to avoid float and use string representations instead. This is your best option, since - according to your function signature - your numbers are represented as strings already.
private function hasMoreThanTwoDecimals(string $number): bool
{
return bcmod(bcmul($number, '100'), '1.0') != 0;
}
This needs the BCMath module to be included in your PHP. A package supporting BCMath and other solutions is brick/math.
The Cheap Solution
However, if you really just need to probe the number and not are doing calculations, you can get the desired result with pattern matching using preg_match.
private function hasMoreThanTwoDecimals(string $number): bool
{
// Trailing 0 does not add to number of decimals
$number = rtrim($number, '0');
return preg_match('~\.\d\d\d~', $number);
}

Answer 2

You can explode the number using the . delimeter, then you return the length of the second part :
$num = 2638655.99;
echo strlen(explode('.',$num)[1]); // Echo 2

Answer 3

Taking the question literally, if a binary floating point number has a maximum of two decimal digits after the decimal point, the fractional part must be one of .0, .25, .5, or .75.
All other binary floating point numbers really have more decimal digits, although printout formatting may hide them. For example, the closest IEEE 754 64-bit binary number to 2638655.99 is 2638655.99000000022351741790771484375, which has more than two digits after the decimal point.
You could subtract the integer part and then test for the remainder being one of the four possibilities.
Alternatively, the real question may be how to determine whether displaying the number will show no more than two digits after the decimal point. If so, convert to string using the appropriate method, then locate the decimal point and count the digits after it, for example as suggested in this answer.

Answer 4

$res = preg_match("^[+-]?([0]{1}|[1-9]{1}[0-9]*)(\.?[0-9]{1,2})?$", $num) == true;
would be the best solution in my opinion. You can use signs (optional) and enforce that a number starts with only one zero.
Possible:
+0.10
+123.01
-1
123
Not possible:
00.0
0001.0
1.
123.123
Be aware that preg_match returns 0 if no match is found and false if an error occurred (preg_match)

Answer 5

you can use the number_format
number_format($number, 2, '.', '');