Is there a way with number_format() to leave out decimal places if the number is not a float/decimal?
For example, I would like the following input/output combos:
50.8 => 50.8
50.23 => 50.23
50.0 => 50
50.00 => 50
50 => 50
Is there a way to do this with just a standard number_format()?
You can add 0 to the formatted string. It will remove trailing zeros.
echo number_format(3.0, 1, ".", "") + 0; // 3
A Better Solution: The above solution fails to work for specific locales. So in that case, you can just type cast the number to float data type. Note: You might loose precision after type casting to float, bigger the number, more the chances of truncating the number.
echo (float) 3.0; // 3
Ultimate Solution: The only safe way is to use regex:
echo preg_replace("/\.?0+$/", "", 3.0); // 3
echo preg_replace("/\d+\.?\d*(\.?0+)/", "", 3.0); // 3
\str_contains($number, '.') ? \rtrim(\rtrim($number, '0'), '.') : $number;
Use:
$a = 50.00;
$a = round($a, 2);
Even though the number has 2 zeros trailing it, if you round it, it won't show the decimal places, unless they have some kind of value.
So 50.00 rounded using 2 places will be 50, BUT 50.23 will be 50.23.
Unless you specify at which point to round up or down, it won't change your decimal values. So just use default round()
If you want to use whitespace here is better solution
function real_num ($num, $float)
{
if (!is_numeric($num) OR is_nan($num) ) return 0;
$r = number_format($num, $float, '.', ' ');
if (false !== strpos($r, '.'))
$r = rtrim(rtrim($r, '0'), '.');
return $r;
}
This function works with PHP 8 and above, or requires a polyfill for the str_contains() function.
function sanitize_float($value): string
{
$value = (string) $value;
if (str_contains($value, '.')) {
$value = rtrim(rtrim((string) $value, '0'), '.');
}
return $value;
}
