Remove leading and trailing slash / in python

Question

I am using request.path to return the current URL in Django, and it is returning /get/category.

I need it as get/category (without leading and trailing slash).

How can I do this?

This is used as ***an example*** in [a meta question](https://meta.stackoverflow.com/questions/420159/poor-sorting-of-answers). That is, not the subject of. — Peter Mortensen, Sep 01 '22 at 17:11

score 253 · Accepted Answer · answered May 02 '12 at 06:37

253

>>> "/get/category".strip("/")
'get/category'

strip() is the proper way to do this.

answered May 02 '12 at 06:37

Amber

9

strip() will also take care of multiple "/"'s – John La Rooy May 02 '12 at 06:41
117

Note also the existence of `lstrip()` and `rstrip()`, in case you want to e.g. remove trailing slashes but preserve leading ones. – Mark Amery Jul 05 '14 at 22:50
18

Actually, I think you should use `strip(os.sep)` – Edward Falk Sep 02 '16 at 23:16
22

For filesystem paths, `os.path.normpath(*path*)` will strip trailing slashes. – Åsmund Jan 18 '17 at 10:31
2

@EdwardFalk that would break if the server is Windows. The context here is URLs. – CrackerJack9 Dec 28 '20 at 18:06

score 18 · Answer 2 · edited May 29 '20 at 10:32

18

def remove_lead_and_trail_slash(s):
    if s.startswith('/'):
        s = s[1:]
    if s.endswith('/'):
        s = s[:-1]
    return s

Unlike str.strip(), this is guaranteed to remove at most one of the slashes on each side.

edited May 29 '20 at 10:32

Alex Tartan

answered May 02 '12 at 06:49

Raymond Hettinger

score 8 · Answer 3 · answered May 02 '12 at 06:56

8

Another one with regular expressions:

>>> import re
>>> s = "/get/category"
>>> re.sub("^/|/$", "", s)
'get/category'

answered May 02 '12 at 06:56

Tim Pietzcker

3 Answers3