查找每个单词中给定/特定字符的首次出现

Question

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.

Answer 1

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

Answer 2

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. Ie \\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.