Python; parse version number with dots from string via regex

Question

So I have this king of string:

some_string-1.4.2.4-RELEASE.some_extension

And I want to parse the version number (in my example: 1.4.2.4) But the number between the dots will not always be 1 digit, it could be something like: 1.40.2.4 or 11.4.2.4.

This is what i have tried:

(\d+\.)?\d+\.\d+

And this does not parse all the numbers.

EDIT

I tried to use the answer from the duplicate link: \d+(\.\d+)+

And according to regex101 I get this result:

Full match  17-24   1.4.2.4
Group 1.    22-24   .4

But in my code I got only .4:

file_name = 'some_string-1.4.2.4-RELEASE.some_extension'

match = re.findall('\d+(\.\d+)+', file_name)
if len(match) == 0:
    print('Failed to match version number')
else:
    print(match[0])
    return match[0]

Sebastien D · Accepted Answer · 2019-06-14 09:37:27Z

0

You might want to consider the following pattern :

file_name = 'some_string-1.4.2.4-RELEASE.some_extension'

pattern = r'\-([0-9.-]*)\-'
match = re.findall(pattern, file_name)
if len(match) == 0:
    print('Failed to match version number')
else:
    print(match[0])

output:

1.4.2.4

answered Jun 14, 2019 at 9:37

Sebastien D

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monK_ · Accepted Answer · 2019-06-14 09:44:23Z

0

Your pattern is almost right.

Use

(\d+(?:\.\d+)+)

This changes the first group to be the entire version number, and ignore the internal repeating group.

str = "some_string-1.4.2.4-RELEASE.some_extension"
regex = r'''(\d+(?:\.\d+)*)'''
print(re.findall(regex, str))  # prints ['1.4.2.4']

answered Jun 14, 2019 at 9:44

monK_

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The fourth bird · Accepted Answer · 2019-06-15 12:04:18Z

0

The pattern \d+(\.\d+)+ contains a repeating capturing group and will contain the value of the last iteration which is .4 and will be returned by findall.

If you would make it a non capturing group it will match the whole value but also values like 1.1 and 9.9.9.99.9.9

\d+(?:\.\d+)+

If the digits must consists of 3 dots and between hyphens, you might use a capturing group:

-(\d+(?:\.\d+){3})-

Regex demo

Or use lookarounds to get a match without using a group:

(?<=-)\d+(?:\.\d+){3}(?=-)

edited Jun 15, 2019 at 12:04

answered Jun 15, 2019 at 11:44

The fourth bird

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