CODEWITHSUNDEEP
javascriptregex
Andre Elrico
Andre Elrico

Reputation: 11480

Finding the first occurrence of a given/ specific character inside every word

stringi = "fuunnynsdfdn dfdfomdfdnd dfdfnntr ndfdf thatnfdfdfdfd"

I was able to make a regEx to DYNAMICALLY find every last occurrence of a character X in every word: (I call it dynamically because I only have to change ONE spot)

(n)(?!(\w+?)?\1(\w+?)?) gives

"fuunnynsdfdn dfdfomdfdnd dfdfnntr ndfdf thatnfdfdfdfd"
            ^          ^       ^   ^         ^

(d)(?!(\w+?)?\1(\w+?)?) gives

"fuunnynsdfdn dfdfomdfdnd dfdfnntr ndfdf thatnfdfdfdfd"
           ^            ^   ^         ^              ^

Question:

How can I get a "dynamic" regEx. That means I have to replace one spot like above to give me:

some regEx with >>> n <<< gives

"fuunnynsdfdn dfdfomdfdnd dfdfnntr ndfdf thatnfdfdfdfd"
    ^                  ^      ^    ^         ^

some regEx with >>> d <<< gives

"fuunnynsdfdn dfdfomdfdnd dfdfnntr ndfdf thatnfdfdfdfd"
         ^    ^           ^         ^          ^

Is it possible to match those with a regEx and that I only have to change ONE spot when I want to match another character?

Please note:

Pure regEx, no js string splitting etc..

The Problem

With the lookbehind I run into a fixed length problem.

Upvotes: 0

Views: 101

Answers (2)

revo
revo

Reputation: 48711

That means I have to replace one spot

You first match n then every thing else:

n(\S*)


var str = "fuunnynsdfdn dfdfomdfdnd dfdfnntr ndfdf thatnfdfdfdfd";
var char = 'n';
console.log(str.replace(new RegExp(char + '(\\S*)', 'g'), '*$1'));
var char = 'd';
console.log(str.replace(new RegExp(char + '(\\S*)', 'g'), '*$1'));

If it's a matter of interest, since Chrome has implemented lookbehinds (that luckily are infnite lookbehinds) you can acheive same result with this regex:

(?<!n\S*)n

Live demo

Upvotes: 2

Tamas Rev
Tamas Rev

Reputation: 7166

In this regex demo, the lookbehind seems to be okay with not fixed length - with javascript. So this is the regex:

(?<=\b[^\Wd]*)d

Explanation:

  • \b is a word barrier
  • [^d]*d finds non-d characters, and then a d
  • [^\Wd] is not a d but word characters. I.e. \W is not a word character, so [^\W] is a word character. So [^\Wd] is every word character, except d
  • \b[^\Wd]*d would match the start of every word, up to the first d. Using the right constructs, this could be enough.
  • (?<=...) is the lookbehind.

Upvotes: 1

Related Questions