python 3 remove before and after on string

Question

I have this string /1B5DB40?full and I want to convert it to 1B5DB40.

I need to remove the ?full and the front /

My site won't always have ?full at the end so I need something that will still work even if the ?full is not there.

Thanks and hopefully this isn't too confusing to get some help :)

EDIT:
I know I could slice at 0 and 8 or whatever, but the 1B5DB40 could be longer or shorter. For example it could be /1B5DB4000?full or /1B5

http://stackoverflow.com/questions/509211/explain-pythons-slice-notation — pvg, Dec 21 '15 at 13:14
@pvg I updated my question. The string will be different sizes, which is what I cannot figure out. If I slice it at 0 and 8 then it wont work — Brandy, Dec 21 '15 at 13:18
@falsetru Will this work even if there isnt `?full` since sometimes the string won't have it? Sorry I'm new to python so trying to figure this all out. — Brandy, Dec 21 '15 at 13:19
Are you actually parsing a part of a URL? because there are library calls that will help you do that — pvg, Dec 21 '15 at 13:20

falsetru · Accepted Answer · 2015-12-21T13:26:47.630

2

Using str.lstrip (to remove leading /) and str.split (to remove optinal part after ?):

>>> '/1B5DB40?full'.lstrip('/').split('?')[0]
'1B5DB40'

>>> '/1B5DB40'.lstrip('/').split('?')[0]
'1B5DB40'

or using urllib.parse.urlparse:

>>> import urllib.parse
>>> urllib.parse.urlparse('/1B5DB40?full').path.lstrip('/')
'1B5DB40'
>>> urllib.parse.urlparse('/1B5DB40').path.lstrip('/')
'1B5DB40'

edited Dec 21 '15 at 13:26

answered Dec 21 '15 at 13:19

falsetru

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enrico.bacis · Answer 2 · 2015-12-21T13:27:24.787

2

You can use lstrip and rstrip:

>>> data.lstrip('/').rstrip('?full')
'1B5DB40'

This only works as long as you don't have the characters f, u, l, ?, / in the part that you want to extract.

edited Dec 21 '15 at 13:27

answered Dec 21 '15 at 13:19

enrico.bacis

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What if `data` is `'/1Buuuu?full'`? – falsetru Dec 21 '15 at 13:21
@falsetru: You are right, I had no idea about that! – enrico.bacis Dec 21 '15 at 13:23

score 1 · Answer 3 · answered Dec 21 '15 at 13:28

1

You can use regular expressions:

>>> import re

>>> extract = re.compile('/?(.*?)\?full')
>>> print extract.search('/1B5DB40?full').group(1)
1B5DB40
>>> print extract.search('/1Buuuuu?full').group(1)
1Buuuuu

answered Dec 21 '15 at 13:28

enrico.bacis

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score 1 · Answer 4 · answered Dec 21 '15 at 13:29

1

What about regular expressions?

import re
re.search(r'/(?P<your_site>[^\?]+)', '/1B5DB40?full').group('your_site')

In this case it matches everything that is between '/' and '?', but you can change it to your specific requirements

answered Dec 21 '15 at 13:29

Mr. E

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score 1 · Answer 5 · answered Dec 21 '15 at 13:33

>>> '/1B5DB40?full'split('/')[1].split('?')[0]
'1B5DB40'
>>> '/1B5'split('/')[1].split('?')[0]
'1B5'
>>> '/1B5DB40000?full'split('/')[1].split('?')[0]
'1B5DB40000'

Split will simply return a single element list containing the original string if the separator is not found.

python 3 remove before and after on string

5 Answers5