在正则表达式中不重复字符的匹配字符串

Question

I'm a total newbie of Regex. What I'm trying to do is to check if a numeric value repeats numbers. They can be anywhere in the string: eg.

123456789 -> would return true
987612345 -> true

but:

122345678 -> would return false because it uses two times the number 2.
182345688 -> false

Is it possible to do this with Regex in PHP?

Answer 1

In case you don't want to use regexes with massive recursive backtracking:

$duplicates = count(count_chars($test, 1)) < strlen($test);

Demo

---

Edit:

In case you want to use a regular expression, you only need to find one duplicate and then quit:

$duplicates = preg_match('/(.).*\1/', $test);

Demo

Answer 2

Re-appearing characters will return 1 , eg:

$match = preg_match_all('/(.).*\1/', '121345678', $arr, PREG_PATTERN_ORDER);

Others will return 0 , eg:

$match = preg_match_all('/(.).*\1/', '12345678', $arr, PREG_PATTERN_ORDER);

Therefore (I named it clean as in "non-repeating"):

$clean = $match == 0;

EDIT:
Maybe for explanation: \\1 is a back-reference to the first (and in this case) only pair of () -s. So this regex is matched when a character is found "that was already there before that occurrence".

Answer 3

If they can be anywhere in your string, it's not too easy. I assume it is somehow possible using regexp. But I recommend doing it another way:

• extract the single characters from the string into a array of characters
• sort the array
• check if two adjacent characters are the same

Or any equivalent technique. But I think this problem is a bit too complicated to be solved by regexp in an elegant manner.

Answer 4

As for regex I'm not 100% sure, but you can do this other way:

function hasRepeatingNumbers($number) {
    $numberArray = array_unique(str_split($number));
    if(count($numberArray) != strlen($number)) {
        return true;
    else
        return false;
}

In the example above we're removing any duplicate numbers and comparing the length of each variables. If they're differente then it's because we removed duplicate numbers.

Then you should just need to:

if(hasRepeatingNumbers('123456789'))
    echo "No repeating numbers";
else
    echo "There are repeating numbers";

This should work just as like.

Answer 5

/(\\d)(?=.*\\1)/

Only looks for digits, matches/quits at first duplicate found.
Warning!, this could be slow.

I imagine this would do it
if ( preg_match( '/(\\d)(?=.*\\1)/', "your string", $match) ) ..

This method might cause problems if digits 0-9 are unique and the string is
very long. Theorhetically, it would inspect 10 times the length of the string.

On the other hand, if you have more than 10 digits, there is at least one dup.
So, in a single pass , extract up to the first 11 digits. Then you can either
loop through the (up to 11 digits) array elements, or use a hash if PHP does that.
This is the fastest method, it might be a verbose regex (11 capture buffers) but PCRE can't do variable amount of capture buffers.

Example in Perl (using a hash):

$_ = '12asasdf3456789 4 0 asdf 3';

my @found = /
 ^
  [^\d]*
  (\d) [^\d]*(\d?)[^\d]*(\d?)[^\d]*(\d?)[^\d]*(\d?)[^\d]*
  (\d?)[^\d]*(\d?)[^\d]*(\d?)[^\d]*(\d?)[^\d]*(\d?)[^\d]*
  (\d?)
/x;

for (@found) {
   if ($seen{$_}++) {
      print "Found a duplicate: '$_'\n";
      last;
   }
}

Output:
Found a duplicate: '4'