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

Answer 1

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'

Answer 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.

Answer 3

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

Answer 4

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

Answer 5

>>> '/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.