regix match words starting with dot but exclude some

Question

I need to build regex to match all words starting with . but also white flag words like .well-known .. or some similar..

For now I have build the one that's complete opposite,, this one captures ONLY this one. I tried to find some regex symbol for invert but that doesn't exist I think..

location ~ /^(\.well-known) {
    deny all;
}

thx

score 1 · Accepted Answer · answered Sep 20 '16 at 10:49

1

Here's an expression that works:

^(\.(?!well-known|other-forbidden-words|another-forbidden-word).+)$

Simply change the other words to white-listed words you want, and add more if need be.

answered Sep 20 '16 at 10:49

ThePerplexedOne

2,920
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thanx man,, now it's great :) https://regex101.com/r/mS2wC6/3 this regex stuff rocks – Kresimir Pendic Sep 20 '16 at 11:29
It does indeed. Regex is very powerful but a pain in the bloody arse. – ThePerplexedOne Sep 20 '16 at 11:38

score 0 · Answer 2 · edited May 23 '17 at 11:48

0

hm,,

I guess that I found my answer thanks to:

Invert match with regexp

this is the test that works now:

https://regex101.com/r/mS2wC6/2

and solution is this:

^(\.(?!well-known))

so ok, I think I got this grouping thing now..

hth, k

edited May 23 '17 at 11:48

Community

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1

answered Sep 20 '16 at 10:45

Kresimir Pendic

3,597
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Your expression isn't matching the whole word when it finds one that starts with `.` and isn't white-listed. – ThePerplexedOne Sep 20 '16 at 10:52
The `^(\.(?!well-known))` is the same as `^\.(?!well-known)` - unless you are using Group 1. – Wiktor Stribiżew Sep 20 '16 at 11:22

score 0 · Answer 3 · answered Feb 07 '18 at 07:35

0

well you don't have to add words in OR "|" condition you can simply do

$pattern = "/\.+(?:[a-zA-Z]|-)/";

which accepts anything that starts with . and contains - or alphabets.

answered Feb 07 '18 at 07:35

Junaid Lodhi

1

regix match words starting with dot but exclude some

3 Answers3