CODEWITHSUNDEEP
javascriptregex
Jtaylorapps
Jtaylorapps

Reputation: 5770

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

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.

Upvotes: 0

Views: 232

Answers (3)

Wrikken
Wrikken

Reputation: 70460

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).
  /

Upvotes: 0

jkshah
jkshah

Reputation: 11703

Try following regex:

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

regex101 demo

Upvotes: 1

Derek
Derek

Reputation: 3435

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?

Upvotes: 0

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