Check for current URL, return true even if there are URL arguments

Question

I'm checking for a very specific pattern in URLs so that a set of code is only executed on the correct type of page. Currently, I've got something like:

/^http:\/\/www\.example\.com\/(?:example\/[^\/]+\/?)?$/;

So, it'll return true for example.com and example.com/example/anythinghere/. However, sometimes this website will append arguments such as ?postCount=25 or something to the end of the URL, so you get:

example.com/example/anythinghere/?postCount=25

Because of this, if I throw the current expression into a conditional, it will return false should there be URL arguments. How would I best go about changing the regex expression to allow for an optional URL argument wildcard, so that, if there's a question mark followed by any additional information, it will always return true, and, if it's omitted, it will still return true?

It would need to return true for:

http://www.example.com/?argumentshere

and

http://www.example.com/example/anythinghere/?argumentshere

As well as those same URLs without the extra arguments.

Drop the `?` before last, change `$` to `(\?|$)` (so the end looks like `...[^\/]+\/)?(\?|$)` — Wrikken, Oct 17 '13 at 20:20
Works like a charm! If you'd like to make it an official answer, I'd be happy to extend the proper thanks. — Jtaylorapps, Oct 18 '13 at 17:54

score 1 · Accepted Answer · answered Oct 17 '13 at 20:24

1

Try following regex:

^http:\/\/www\.example\.com(?:\/example\/[^\/]+\/?)?\??.*$

regex101 demo

answered Oct 17 '13 at 20:24

jkshah

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Will give it a try tomorrow and get back to you! – Jtaylorapps Oct 17 '13 at 21:22

score 0 · Answer 2 · edited May 23 '17 at 12:32

0

You can build the URL without parameters and compare that with your current expression.

location.protocol + '//' + location.host + location.pathname

How to get the URL without any parameters in JavaScript?

edited May 23 '17 at 12:32

Community

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answered Oct 17 '13 at 20:19

Derek

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score 0 · Answer 3 · answered Oct 19 '13 at 00:39

Upgrading my comment to an answer:

 /^http:\/\/www\.example\.com\/(?:example\/[^\/]+\/?)?$/;

Meanse:

 /^    # start of string
      http:\/\/www\.example\.com\/  #literal http://www.example.com/
      (?:           
         example\/[^\/]+\/? #followed by example/whatever (optionally closed by /)
      )?
      $ end-of-string
  /

The main problem here, is your requirements ("followed by an optional querystring") does not match you regex (which requires an end-of-string). We solve it by:

 /^    # start of string
      http:\/\/www\.example\.com\/  #literal http://www.example.com/
      (?:           
         example\/[^\/]+\/? #followed by example/whatever (optionally closed by /)
      )?
      (\?|$) followed by either an end-of-string (original), or a literal `?` (which in url context means the rest is a query string and not a path anymore).
  /

Check for current URL, return true even if there are URL arguments

3 Answers3