Regex: How to get all strings between slashes from URL pathname?

DevonDahon 0
Viewed 47

How to get all strings between slashes from URL pathname until first interrogation mark if any using regex?

For instance, I can have:

/abc/def/ghi?foo=bar → ['abc','def','ghi']
/abc/xyz             → ['abc','xyz']
abc/xyz              → ['abc','xyz']
abc/xyz/             → ['abc','xyz']

I tried this javascript code:

'/abc/def/ghi?foo=bar'.match(/\w+/)

But I'm only getting the abc.

regex javascript
2 Answers

Converting my comment to answer so that solution is easy to find for future visitors.

You may use this regex with look around assertions:

/(?<=^|\/)[^\/?]+(?=[\/?]|$)/gm

RegEx Demo

RegEx Breakup:

  • (?<=^|\/): Lookbehind to assert presence of line start or / at previous position
  • [^\/?]+: Match 1+ of any char that is not / and ?
  • (?=[\/?]|$): Lookahead to assert presence of line end or / or ?` at next position
anubhava0

An alternative without a lookbehind, using an alternation and a capture group checking the value of group 1 in the callback of Array.from

If there is a group value, then return it, else return the match.

\/(\w+)|^\w+(?=\/)

Regex demo

const regex = /\/(\w+)|^\w+(?=\/)/g;
[
  "/abc/def/ghi?foo=bar",
  "/abc/xyz",
  "abc/xyz",
  "abc/xyz/",
  "abc",
].forEach(s => console.log(
  Array.from(s.matchAll(regex), m => m[1] ? m[1] : m[0])));

Or excluding the / and the ?

const regex = /\/([^\/?]+)|^[^\/?]+(?=\/)/g;
[
  "/abc/def/ghi?foo=bar",
  "/abc/xyz",
  "abc/xyz",
  "abc/xyz/",
  "abc",
].forEach(s => console.log(
  Array.from(s.matchAll(regex), m => m[1] ? m[1] : m[0])));

The fourth bird0
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