How to extract path from URL using Regex?

Question

I want to extract path from url.

www.abc.com/contact/abc => /contact 
www.abc.com/contact.abc/asdsad => /contact.abc
www.abc.com/contact-abc.html/asdsad => /contact-abc.html
www.abc.com/contact-2 => /contact-2
www.abc.com/contact-us => /contact-us
abc&/contact.abc&^ = /contact.abc

What single regex could do this?

You could use split at / and get second entry in the array insteed and in abc&/contact.abc&^ you can replace not wished parts. — asdf
– asdf, Commented Oct 15, 2016 at 11:35

Cristofor · Accepted Answer · 2016-10-15 13:16:45Z

4

\/([0-9a-zA-Z+.-]+)[\/&| ]

This one could be a solution, though it is strictly related to the urls you gave in your question as examples.

I'll leave you the link to regexr so that you can understand what is there actually behind.

Regexr resource

answered Oct 15, 2016 at 13:16

Cristofor

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1 Comment

Mai Hữu Lợi Over a year ago

But It needs to be exactly prefix with /contact. Thank you.

Community · Accepted Answer · 2017-05-23 12:23:04Z

2

If you are using Java, use java.net.URL class to parse URL and getPath method to get its path as explained here, no need for regex:

https://docs.oracle.com/javase/tutorial/networking/urls/urlInfo.html.

For regex, refer to this in question:

Getting parts of a URL (Regex)

edited May 23, 2017 at 12:23

CommunityBot

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answered Oct 15, 2016 at 11:36

Bambang

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