parsing a path based on regex - Python

Question

I'm trying to scan an Amazon S3 bucket, in order to search if a new version of our installer has been posted:

The bucket scan returns me something like:

versions/
versions/4.4.1.2/
versions/4.4.1.2/Installer.sh
versions/4.4.2.11/
versions/4.4.2.11/Installer.sh
versions/4.5.0.10a/
versions/4.5.0.10a/Installer.sh
versions/4.5.0.12a/
versions/4.5.0.12a/Installer.sh

I only need to get the <{d}.{d}.{d}.{d}{a-z]}> part between versions ... how do I parse that using regex in Python?

Answer 1

Use re module:

import re
text = "versions/4.5.0.10a/"
re.search('\d+\.\d+\.\d+\.\d+[a-z]', text).group()
Output: '4.5.0.10a'

Answer 2

("versions/4.4.1.2/").split("/")[1]

Answer 3

You could do this through Positive lookahead and lookbehind.

>>> s = "versions/";
>>> m = re.compile(r'(?<=/).*?(?=/)')
>>> re.search(m, s)
>>> s = "versions/4.4.1.2/";
>>> re.search(m, s).group()
'4.4.1.2'
>>> s = "versions/4.4.1.2/Installer.sh";
>>> re.search(m, s).group()
'4.4.1.2'