Regex unable to parse version

Question

My objective is to get the version number from a string where the string may contain anything. Here's my example data set:

Version 1.32.0.1
Version 1.32.0.1c
Version 1.32.1
Version 1.33.2e
Version 1.32

I've attempted to match that with this regex (\d+\.\d+(?:\.\d+)?)(\w?) but I can't seem to figure out why this won't match the 4th decimal value, even with the descriptive breakdown provided by regex101.com.

What am I misunderstanding about this regex causing it to not match all version variations?

In .NET you can do the following: string version = Regex(str_version, @"Version\s+(?'ver'.+)").Groups["ver"].Value; — JohnyL
– JohnyL, Commented Jan 12, 2019 at 16:22

mukesh210 · Accepted Answer · 2019-01-12 14:51:56Z

1

Currently you have used ? in non-capturing group which will match only 3rd decimal value(even if you have 3rd and 4th decimal value present). One solution would be to use * instead of ? in non-capturing group for your non-capturing group to match 3rd as well as 4th decimal value.

I tried in Scala and it works perfectly for your inputs above:

  val text: String = "Version 1.32.0.1c"
  val pattern = """(\d+\.\d+(?:\.\d+)*)(\w?)""".r
  val res: Regex.MatchIterator = pattern.findAllIn(text)
  println(res.group(1)) // 1.32.0.1
  println(res.group(2)) // c

answered Jan 12, 2019 at 14:51

mukesh210

2,8922 gold badges23 silver badges42 bronze badges

Sign up to request clarification or add additional context in comments.

Comments

Sweeper · Accepted Answer · 2019-01-12 16:25:36Z

1

Count how many \d+ you have got. You should count 3. So your regex will at most match 3 numbers, and it cannot possibly match 4.

I'm not sure if this is a typo or you genuinely don't understand, but this can be fixed by adding another group:

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

This can be shortened to:

\d+(?:\.\d+){1,3}(\w?)

If you want to match any number of numbers more than 1, you can do:

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

\w would match _ as well. If you don't want that, you can replace it with [a-z].

edited Jan 12, 2019 at 16:25

answered Jan 12, 2019 at 14:01

Sweeper

292k23 gold badges261 silver badges439 bronze badges

2 Comments

Andreas Over a year ago

You can modify the short version to cover the long version, i.e. max 4 numbers, by using {1,3}, as in: \d+(?:\.\d+){1,3}[a-z]?. I change to use [a-z] since OP probably doesn't want to match _, which \w would.

Sweeper Over a year ago

@Andreas Thanks for the comments. Edited.

asdator1213 · Accepted Answer · 2019-01-13 11:43:17Z

0

Try this regex, maybe it's the result you want to achieve:

\w+[.](\w*[.]?)*

In a text block:

Version 1_32.0.1
Version 1.32.0.1c
Version 1.32.1
Version 1.33.2e.
Version 1.32
Version e342

it matches only version numbers which starts with any alphanumeric characters (and underscores), followed by one dot and in the loop followed by zero or many any of alphanumeric character with dot at the end of a line.

Here is a demo: regex

edited Jan 13, 2019 at 11:43

answered Jan 13, 2019 at 11:17

asdator1213

273 bronze badges

Collectives™ on Stack Overflow

Regex unable to parse version

3 Answers 3

Comments

2 Comments

Comments

Your Answer

Hot Network Questions

Collectives™ on Stack Overflow

3 Answers 3

Comments

2 Comments

Comments

Your Answer

Sign up or log in

Post as a guest

Related