Looking for regular expression that matches the first part of a URL fragment

Question

I have a URL fragment like 'users/123/edit' and I would like to match the first part of the string (in this case 'users'). The URL fragment could also be just 'users' and it should also match 'users'.

I could do it via JS find() and substr(), but I would like to see, how I do it via a Regular Expression. An explanation to the solution would be nice aswell.

score 2 · Answer 1 · answered Mar 27 '14 at 11:32

2

Regex: /([a-z0-9]+)(\/[a-z0-9]+)*/i

Matches:

users/123/edit
users
myuser123 // tell if not required

Explanation:

([a-z0-9]+) catches one or more occurrences (+) of [a-z0-9] and captures it.

(\/[a-z0-9]+)* catches zero or more occurrences (*) of / followed by [a-z0-9] and captures the last occurrence.

The ignore case flag makes the entire match case-insensitive.

DEMO

answered Mar 27 '14 at 11:32

Gaurang Tandon

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I accepted radia's answer, because it only matched the required base-part. Still, thanks for the good explanation! – johnny Mar 27 '14 at 12:17
@johnny No problem at all :) – Gaurang Tandon Mar 27 '14 at 12:19

score 1 · Answer 2 · answered Mar 27 '14 at 11:32

1

You can use split('/')[0]:

s = 'users/123/edit'.split('/')[0];
// users

OR else:

s = 'users'.split('/')[0];
// users

answered Mar 27 '14 at 11:32

anubhava

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radia · Accepted Answer · 2014-03-27T11:38:10.837

1

using regex :

str.match(/[^\/]*/i)[0]

this matches all characters different then (before) "/"

FIDDLE

edited Mar 27 '14 at 11:38

answered Mar 27 '14 at 11:33

radia

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Looking for regular expression that matches the first part of a URL fragment

3 Answers3