RegEx to handle Alphanumeric Input - PHP

I have this code.

<?php
    $USClass = "3/312";
    $USClass = preg_replace_callback('~[\d.]+/[\d.]+~', function ($matches) {
    $parts = explode('/', $matches[0]);
    return $parts[1] . ',' . $parts[0];
    }, $USClass);
    echo $USClass;
?>

It prints 312,3 which is what I wanted.

However, if I give an input like D12/336 then it does not work. I want it to print 336,D12

How can I do it? and What is wrong with my current code which is not handling this Alphanumeric? Is it because I used \d ?

EDIT:

I want it to handle inputs like t开发者_StackOverflow社区his as well

148/DIG.111

then the output should be DIG.111,148

---

Yes \d does only contain digits.

You can try \w instead this is alphanumeric, but additionally it includes also _

To be Unicode you can go for

~[\pL\d.]+/[\pL\d.]+~u

\pL is a Unicode code point with the property "Letter"

The u at the end turn on the UTF-8 mode that is needed to use this feature

See http://www.php.net/manual/en/regexp.reference.unicode.php

Other solution

I think you ar e doing this a bit complicated. It would be simplier if you would make use of capturing groups.

Try this:

$in = "148/DIG.111";
preg_match_all('~([\w.]+)/([\w.]+)~', $in, $matches);
echo $matches[2][0] . ',' . $matches[1][0];

Explanation:

([\w.]+)/([\w.]+)
^^^^^^^^ ^^^^^^^^
Group 1   Group 2

Because of the brackets the matched substring is stored in the array $matches.

See here for more details on preg_match_all

---

With a simple preg_replace:

$USClass = "148/DIG.111";
$USClass = preg_replace('#(.+?)/(.+)#', "$2,$1", $USClass);
echo $USClass;

output:

DIG.111,148