Regular expression exclude double character

Question

I need to have regular expression which find a character which not following by same character after it. Mean exclude double even multiple same character.

For exampl

Answer 1

UPDATE:

(?<!e)e(?!e)

Will match e not with negative lookbehind to prevent preceeding e and negative lookahead preventing following e.

Can be checked here

\b(([A-Za-z])(?!\2))+\b

Will match a word (sequence of one or more characters between A-Za-Z), with negative lookahead which prevents following character to be the same as last match, group 2, or using non capturing group.

/\b(?:([A-Za-z])(?!\1))+\b/g

however only will match because it doesn't contain repeated character. to match a word containing e but no ee

/(?<![a-z])(?=[a-z]*e)(?![a-z]*ee)[a-z]+/gi

Answer 2

/\b([a-df-z]*e[a-df-z]*)\b\s*/g

You could add the flag case insensitive /i if needed.

Explanation:

/               : regex delimiter
  \b            : word boundary
  (             : start group 1
    [a-df-z]*   : 0 or more letter that is not "e"
    e           : 1 letter "e"
    [a-df-z]*   : 0 or more letter that is not "e"
  )             : end group 1
  \b            : word boundary
  \s*           : 0 or more spaces
/g              : regex delimiter, global flag

As you didn't give which language you're using, here is a perl script:

my $str = "need only single characteeer";
my @list = $str =~  /\b([a-df-z]*e[a-df-z]*)\b\s*/g;
say Dumper\@list;

Output:

$VAR1 = [
          'single'
        ];

And a php script:

$str = "need only single characteeer";
preg_match_all("/\b([a-df-z]*e[a-df-z]*)\b\s*/", $str, $match);
print_r($match);

Output:

Array
(
    [0] => Array
        (
            [1] => single 
        )

    [1] => Array
        (
            [1] => single
        )

)