How can i Select Everything In Url except filename and extension?

Question

https://fire.vimeocdn.com/1485599447-0xf546ac1afe7bce06fa5153973a8b85b1c45051d3/159463108/video/499604330/playlist.m3u8

I want to Include Everything Except playlist.m3u8

(playlist.[^.]*)

is selecting "playlist.m3u8", i need to do exactly opposite.

Here is an Demo. https://regex101.com/r/RONA65/1

Possibly http://stackoverflow.com/questions/24465512/regular-expression-to-match-everything-until-the-last-occurrence-of would be helpful. — Shakir Khan, Jan 28 '17 at 07:21
@Nowhere Check my solution. I think you were looking for positive look ahead. — Mohammad Yusuf, Jan 28 '17 at 07:44

Mohammad Yusuf · Accepted Answer · 2017-01-28T07:43:32.913

1

You can use positive look ahead:

(.*)(?=playlist\.[^.]*)

Demo:

https://regex101.com/r/RONA65/4

Or you can try it like this as well:

.*\/

Demo:

https://regex101.com/r/RONA65/2

Regex:

.*\/ Select everything till last /

edited Jan 28 '17 at 07:43

answered Jan 28 '17 at 07:14

Mohammad Yusuf

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Second Approach is Good, select all till /, Winner – Rahul Sharma Jan 28 '17 at 08:04

score 0 · Answer 2 · answered Jan 28 '17 at 07:16

You can use the split function:

>>> s = 'https://fire.vimeocdn.com/.../159463108/video/499604330/playlist.m3u8'
>>> '/'.join(s.split('/')[:-1])
'https://fire.vimeocdn.com/.../159463108/video/499604330'

Or simpler with rsplit:

>>> s = 'https://fire.vimeocdn.com/.../159463108/video/499604330/playlist.m3u8'
>>> s.rsplit('/', 1)[0]
'https://fire.vimeocdn.com/.../159463108/video/499604330'

score 0 · Answer 3 · answered Jan 28 '17 at 07:23

0

Use non-greedy match by adding '?' after '*'

import re
s = 'https://fire.vimeocdn.com/1485599447-0xf546ac1afe7bce06fa5153973a8b85b1c45051d3/159463108/video/499604330/playlist.m3u8'
m = re.match('(.*?)(playlist.[^.]*)', s)
print(m.group(1))

answered Jan 28 '17 at 07:23

gzc

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How can i Select Everything In Url except filename and extension?

3 Answers3