Finding all the ten different digits in a random string

Question

Sorry if this is answered somewhere, but I couldn't find it.

I need to write a regexp to matches on strings that contain the digits from 0 to 9 exactly once. For example:

e8v5i0l9ny3hw1f24z7q6

You can see that numbers [0-9] are present exactly once and in random order. (Letters are present also exactly once, but that is an advanced quest...) It must not match if a digit is missing or if any digit is present more than one time.

So what would be the best regexp to match on strings like these? I am still learning regex and couldn't find a solution. It is PCRE, running in perl environment, but I cannot use perl, only the regex part of it. Sorry for my english and thank you in advance.

I'm sure that there are much better ways to solve this rather than using regex. — devnull, Apr 14 '14 at 10:10
Do you want one regexp or ten regexps with and operators would be ok too? :) — Maxime Chéramy, Apr 14 '14 at 10:11
What do you mean by sorting? I can not alter the string in any way.:( Maybe one regexp is better than ten with and operators.:) But I am curious of the latter, can you show? — Tom, Apr 14 '14 at 11:30
If you can have a sorted string, you can then look for .*0.*1.*2.*... — Maxime Chéramy, Apr 14 '14 at 12:42
Classic case of using a hammer to crack a nut. There are simple and clear ways to this. Regular expressions aren't one of them. — user207421, Apr 15 '14 at 08:16

score 6 · Accepted Answer · edited May 23 '17 at 12:13

What about this pattern to verify the string:

^\D*(?>(\d)(?!.*\1)\D*){10}$

^\D* Starts with any amount of characters, that are no digit
(?>(\d)(?!.*\1)\D*){10} followed by 10x: a digit (captured in first capturing group), if the captured digit is not ahead, followed by any amount of \D non-digits, using a negative lookahead. So 10x a digit, with itself not ahead consecutive should result in 10 different [0-9].

\d is a shorthand for [0-9], \D is the negation [^0-9]

Test at regex101, Regex FAQ

If you need the digit-string then, just extract the digits, e.g. php (test eval.in)

$str = "e8v5i0l9ny3hw1f24z7q6";

$pattern = '/^\D*(?>(\d)(?!.*\1)\D*){10}$/';

if(preg_match($pattern, $str)) {
  echo preg_replace('/\D+/', "", $str);
}

This seems to work as I think it should work! At least at debuggex.com says so. :) Thank you! — Tom, Apr 14 '14 at 11:22

score 0 · Answer 2 · answered Apr 14 '14 at 10:29

0

It is easy to create a regular expression that matches one specific permutations of the numbers and ingnore the other characters. E.g.

^[^\d]*0[^\d]1[^\d]*2[^\d]*3[^\d]*4[^\d]*5[^\d]*6[^\d]*7[^\d]*8[^\d]*9[^\d]*$

You can combine 10! expressions for every possible permutation with |

Although this is completely inpractical it shows that such a regular expression (without lookahead) is indeed possible.

However this is something that is much better done without regular expression matching.

answered Apr 14 '14 at 10:29

Udo Klein

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you can use lookaheads.You dont need all 10*9*8*7... permutations – Anirudha Apr 14 '14 at 10:30
Yes of course, but as I understand it lookahead is an extension. Strictly speaking this is then not a reuglar expression anymore. – Udo Klein Apr 14 '14 at 10:32
But with this you determine the order of the numbers, so they cannot be in random order. – Tom Apr 14 '14 at 11:24
That's why I wrote that it would require 10! expressions of this type. – Udo Klein Aug 23 '14 at 08:08

score 0 · Answer 3 · answered Apr 15 '14 at 08:12

$s = "e8v5i0l9ny3hw1f24z7q6";
$s = preg_replace('/[^\d]/i', '', $s); //remove non digits
if(strlen($s) == 10) //do we have 10 digits ?
if (!preg_match('/(\d)(\1+)/i', $s))  //if no repeated digits
echo "String has 10 different digits";

http://ideone.com/eY4eGx

Finding all the ten different digits in a random string

3 Answers3