I'm trying to trim the leading and trailing zeros from a string that contains a numeric value used to represent a percentage.
I wrote this ugly block of code that combines rtrim(), ltrim() with a preg_replace():
rtrim(ltrim(preg_replace("/\d*\.(0(?![0$])|[1-9])*\K0+$/", "", $value),"0"), ".")
But it doesn't seem to work since if I enter 0010.003000 the output is 10.003000 instead of the desired 10.003
Is there a way to make a regex read right to left in Php (delete zeros from the right until you find a zero or a [1-9])? Or is there a way to improve my regex to accomplish my goal?
Important note: For some arguable reason I can't do a simple cast to float and simply enjoy the time saved playing videogames.
To better explain my strange question here are some samples with expected behaviors: 01.00 => 1 , 100 => 100 , 002.0050 => 2.005 , 0.005 => .005
Try the following:
/^0+|\..+?\K0+$/g
This either matches one or more 0s at the start-of-string ^0+, or it matches one or more 0s at the end-of-string 0+$.
In PHP:
preg_replace_all("/^0+|\..+?\K0+$/", "", $value)
NOTE: I would recommend using the non regular expression solutions because they are more suitable for this simple situation. Nevertheless I posted this because you asked for a regular expression.
Simply cast your string to float:
$s = '0010.003000';
echo (float) $s;
//=> 10.003
You can use floatval instead of regex:
<?php
$str = "0010.003000";
print floatval($str);
Output: 10.003
echo trim('0010.003000', '0');
What would you say about such regex?
echo preg_replace("/((^0+)|(0+$))/", "", $value);
Ok, now I understand the problem ;) Try this:
echo preg_replace("/^0*([1-9][0-9]*(\.[0-9]*[1-9]))?0*$/", "$1", $value);
To answer your strange question, I added support for 0.000 that returns 0:
$nbr = preg_replace('/^(0)+\.?0*$|^0+|\.0*$|\.\d+?\K0+$/', '$1', $nbr);