Express.JS get url path without any additional parameters

Question

Is there any straightforward way to get URL path segment without any additional data (url parameters, css '#', etc.)

for example:

full URL: http://example.com/panel/list
path segment I want to get: /panel/list

full URL: http://example.com/panel/list?user=joe
path segment I want to get: /panel/list

full URL: http://example.com/panel/list#header
path segment I want to get: /panel/list

And any other combination of data that is not path segment of my URL

Does this answer your question? node.js with express how to remove the query string from the url — Jakub Kukul
– Jakub Kukul, Commented Jun 5, 2021 at 11:49

rkm · Accepted Answer · 2017-03-11 16:43:39Z

10

I think you can use req.baseUrl + req.path.

And for full URL use req.originalUrl.

answered Mar 11, 2017 at 16:43

rkm

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2 Comments

Jakub Pastuszuk Over a year ago

but if I type /admin I get /admin/ instead

Jakub Pastuszuk Over a year ago

Edit, I just used replace to trim last trailing slash (req.baseUrl + req.path).replace(/\/$/, "")

Vishnu Jayan · Accepted Answer · 2018-11-09 12:06:11Z

1

For full url without any parameters, you can use req.protocol + '://' + req.get('host') + req.baseUrl + req.path;

answered Nov 9, 2018 at 12:06

Vishnu Jayan

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Comments

Tyler2P · Accepted Answer · 2024-04-24 17:42:39Z

0

In frontend you can use window.location.pathname.

edited Apr 24, 2024 at 17:42

Tyler2P

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answered Apr 23, 2024 at 6:24

Mumin Ahmad Khan

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