Levenshtein distance in regular expression

is there possiblity how to include levenshtein distance in regular expression query?

No, not in a sane way. Implementing - or using an existing - Levenshtein distance algorithm is the way to go.

There are a couple of regex dialects out there with an approximate matching feature - namely the TRE library and the regex PyPI module for Python.

The TRE approximate matching syntax is described in the "Approximate matching settings" section at https://laurikari.net/tre/documentation/regex-syntax/. A TRE regex to match stuff within Levenshtein distance 1 of hello would be:

(hello){~1}

The regex module's approximate matching syntax is described at https://pypi.org/project/regex/ in the bullet point that begins with the text Approximate “fuzzy” matching. A regex regex to match stuff within Levenshtein distance 1 of hello would be:

(hello){e<=1}

Perhaps one or the other of these syntaxes will in time be adopted by other regex implementations, but at present I only know of these two.

You can generate the regex programmatically. I will leave that as an exercise for the reader, but for the output of this hypothetical function (given an input of "word") you want something like this string:

"^(?>word|wodr|wrod|owrd|word.|wor.d|wo.rd|w.ord|.word|wor.?|wo.?d|w.?rd|.?ord)$"

In English, first you try to match on the word itself, then on every possible single transposition, then on every possible single insertion, then on every possible single omission or substitution (can be done simultaneously).

The length of that string, given a word of length n, is linear (and notably not exponential) with n.

Which is reasonable, I think.

You pass this to your regex generator (like in Ruby it would be Regexp.new(str)) and bam, you got a matcher for ANY word with a Damerau-Levenshtein distance of 1 from a given word.

(Damerau-Levenshtein distances of 2 are far more complicated.)

Note use of the (?> non-backtracing construct which means the order of the individual |'d expressions in that output matter.

I could not think of a way to "compact" that expression.

EDIT: I got it to work, at least in Elixir! https://github.com/pmarreck/elixir-snippets/blob/master/damerau_levenshtein_distance_1.exs

I wouldn't necessarily recommend this though (except for educational purposes) since it will only get you to distances of 1; a legit D-L library will let you compute distances > 1. Although since this is regex, it would probably work pretty fast once constructed (note that you should save the "compiled" regex somewhere since this code currently reconstructs it on EVERY comparison!)