Regular expression currency format with dots and comma - php

Question

My goal is getting something like that: 150.000,54 or 48.876,05 which means my commas are decimal starters.
Here's my code so far :
<?php
//cut numbers after comma if there are any, after 2 digits
$matchPattern = '/[0-9]+(?:\,[0-9]{2}){0,2}/';
//remove everything except numbers, commas and dots
$repl1 = preg_replace("/[^a-zA-Z0-9,.]/", "", $input);
//let there be a 0 before comma to have values like 0,75
$repl2 = preg_replace("/^[0]{1}$/", "",$repl1);
//now i need you here to help me for the expression putting dots after each 3 numbers, until the comma:
$repl3 = preg_replace("/regexphere$/", ".", $repl2);
preg_match($matchPattern, $repl3, $matches);
echo($matches[0]);
?>
I know preg_replacing 3 times is stupid but I am not good at writing regular expressions. If you have a better idea, don't just share it but also explain. I know a little of the types : http://regexone.com/lesson/0
Thank you in advance.
--------UPDATE--------
So I need to handle 0000,45 like inputs to 0,45 and like 010101,84 inputs to 1,84
When this is done, I'm done.
$input = Input::get('userinput');
$repl1 = preg_replace("/[^0-9,.]/", "", $input);
$repl2 = preg_replace("/^0/", "",$repl1);
$repl3 = str_replace(".","",$repl2);
preg_match('/[0-9]+(?:\,[0-9]{2}){0,2}/', $repl3, $matches);
$repl4 = preg_replace('/(\d)(?=(\d{3})+(?!\d))/', '$1.', $matches[0]);
return repl4;
----UPDATE----
Here's what i get so far : https://ideone.com/5qmslB
I just need to remove the zeroes before the comma, before the numbers.

Answer 1

I am not sure this is the best way, but I hope it is helpful.
Here is the updated code that I used with a fake $input:
<?php
$input = "textmdwrhfejhg../,2222233333,34erw.re.ty";
//cut numbers after comma if there are any, after 2 digits
$matchPattern = '/[0-9]+(?:\,[0-9]{2}){0,2}/';
//remove everything except numbers, commas and dots
$repl1 = trim(preg_replace("/[^0-9,.]/", "", $input), ".,");
echo "$repl1" . "\n";
//let there be a 0 before comma to have values like 0,75, remove the 0
$repl2 = preg_replace("/^0/", "",$repl1);
echo "$repl2" . "\n";
//The expression putting dots after each 3 numbers, until the comma:
$repl3 = preg_replace('/(\d)(?=(?:\d{3})+(?!\d))/', '$1.', $repl2);
echo "$repl3" . "\n";
The expression putting dots after each 3 numbers is
(\d)(?=(?:\d{3})+(?!\d))
Here, you can see how it works. In plain human,
(\d) - A capturing group that we'll use in the replacement pattern, matching a single digit that....
(?=(?:\d{3})+(?!\d)) - is followed by groups of 3 digits. External (?=...) is a look-ahead construction that checks but does not consume characters, (?:\d{3})+ is a non-capturing group (no need to keep the matched text in memory) that matches 3 digits exactly (due to the limiting quantifier {3}) 1 or more times (due to the + quantifier), and (?!\d) is a negative look-ahead checking that the next character after the last matched 3-digit group is not a digit.
This does not work in case we have more than 3 digits after a decimal separator. With regex, I can only think of a way to support 4 digits after decimal with (?<!,)(\d)(?=(?:\d{3})+(?!\d)). Not sure if there is a generic way without variable-width look-behind in PHP (as here, we also need a variable-width look-ahead, too). Thus, you might consider splitting the $repl2 value by comma, and only pass the first part to the regex. Then, combine. Something like this:
$spl = split(',', $repl2); // $repl2 is 1234,123456
$repl3 = preg_replace('/(\d)(?=(?:\d{3})+(?!\d))/', '$1.', $spl[0]);
$repl3 .= "," . $spl[1]; // "1.234" + "," + "123456"
echo "$repl3" . "\n"; // 1.234,123456
Update:
The final code I have come up with:
$input = "textmdwrhfejhg../0005456,2222233333,34erw.re.ty";
//Here's mine :
$repl1 = trim(preg_replace("/[^0-9,.]/", "", $input), '.,');
//following line just removes one zero, i want it to remove all chars like
//Input : 000549,569 Output : 549,569
echo "$repl1\n";
$repl2 = preg_replace("/^0+(?!,)/", "",$repl1);
$repl3 = str_replace(".","",$repl2);
preg_match('/[0-9]+(?:\,[0-9]{2}){0,2}/', $repl3, $matches);
$repl4 = preg_replace('/(\d)(?=(\d{3})+(?!\d))/', '$1.', $matches[0]);
echo $repl4;