regex extract last occurrence ignoring new line

Question

Considering a string like:

 "@foo #foo@foo@foo #foo\n
  foofoofoo\n 
  foo @bar"

I try for 2 days to extract the last #/@ occurrence so here, the @ before 'bar'. For now, i have something like this [@#](?!.*[@#]) which seems to work except when user insert new lines in there.

Can someone give me some tips please?

`.` does not match new line by default, do a catch all like `[\s\S]` — Paul S., Dec 02 '15 at 15:02
Related: http://stackoverflow.com/questions/1979884/how-to-use-javascript-regex-over-multiple-lines — Oli, Dec 02 '15 at 15:03

score 0 · Accepted Answer · answered Dec 02 '15 at 15:04

0

You can use this lookhahead regex:

/[#@](?![\s\S]*[#@])/

RegEx Demo

(?![\s\S]*[#@]) is the negative lookahead that asserts there is no @ or # ahead of current position in any line. [\s\S] matches any character including newline.

answered Dec 02 '15 at 15:04

anubhava

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Ty for your help but for a string like '@foo #bar' this regex gives me '@' when i expect '#' – nantes_flo Dec 02 '15 at 15:28
`@foo #bar` will indeed match `#` [as you can see in this demo](https://regex101.com/r/gS5qQ1/2) – anubhava Dec 02 '15 at 15:31
1

You were obvisouly right ty, helps me a lot ! – nantes_flo Dec 02 '15 at 15:43

regex extract last occurrence ignoring new line

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