Regex: Matching one substring, but excluding another substring later on

Question

Trying to write a regex that matches certain URLs. For our purposes, let's assume the URLs are:

http://website.com/Section/subsection/Cows
http://website.com/Section/Cows
http://website.com/Section/subsection/Chickens

I want to match:

URLs that contain /Section/
Unless it is followed by /Chickens

The closest I've gotten is /\/Section\/([a-zA-Z0-9]+)\/(?!Chickens)/gi

This works for the first URL, but the 2nd URL will not be matched. I know it is because it doesn't have the [a-zA-z0-9]+ section, but I don't know how to solve it.

url.indexOf('/Section') != -1 && url.indexOf('/Chickens') != -1 — adeneo
– adeneo, Commented Dec 19, 2013 at 19:40
Do you need to make sure that its a valid URL? If so, you will need a much larger regex. I would use someone else's URL matcher code and then do what @adeneo suggests. — aebabis
– aebabis, Commented Dec 19, 2013 at 20:14

Jon · Accepted Answer · 2013-12-19 19:47:42Z

3

You were quite close. Here is the regex:

\/Section\/(?!.*\/Chickens)

It just matches the section part and then asserts that anything followed by "/Chickens" cannot match going forward.

You can tail the regex off with an additional .* (outside the negative lookahead) if you want it to capture the URL path instead of just testing for a match.

answered Dec 19, 2013 at 19:47

Jon

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anubhava · Accepted Answer · 2013-12-19 19:48:54Z

0

You're close. Try this regex:

/\/Section\/([a-z0-9]+)(?!\/Chickens)/gi

answered Dec 19, 2013 at 19:48

anubhava

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score 0 · Accepted Answer · 2013-12-20 09:41:48Z

0

This one rejects "/Section/Chickens" as well :

var r = /\/Section\/(?!(.+\/)*Chickens(\/|\?|#|$))/;
r.test('/Section/Cows'); // true
r.test('/Section/Chickens'); // false
r.test('/Section/CowsChickens'); // true
r.test('/Section/ChickensCows'); // true
r.test('/Section/Cows/Chickens'); // false
r.test('/Section/Cows/Cows/Chickens'); // false
r.test('/Section/Cows/Chickens/Cows'); // false

edited Dec 20, 2013 at 9:41

answered Dec 20, 2013 at 9:24

user1636522

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