Match version numbers using regex while excluding entries with underscore

user_3008444 0
Viewed 85

I'm trying to match release versions using regex but exclude any string or group in a string that might have underscore after version number. The pattern is n.n(optionally additional .n) and nothing else, aka major.minor version and additionally build version. Update: Additionally if possible, letters can be allowed but only after n.n or n.n.n for example n.na or n.n.na but not any other combination.

For example:

1.2.3
10.2.4
10.20.5
10.20.323
1.20.30
1.2.33
1.0
1.2a
1.1.2a
1.22_UAT_2
1.10.2_TEST1
1.2_UAT2
2.0_LIVE_2_
a line with text which has 1.2 or then 1.2.1 but then 1.2.1_UAT2 or 1.3_TEST also include version as 1.3a or 1.4.1b but not 1.2a_UAT3 and not 1.2b_1 or 1.2.3a_b

Should return:

1.2.3
10.2.4
10.20.5
10.20.323
1.20.30
1.2.33
1.0
1.2a
1.1.2a
1.2
1.2.1
1.3a
1.4.1b
regex
4 Answers

Using ^([\d\.]*\n):

( ) marks a group

^ is used to search at the start of a Line

[\d\.]* matches any digit or ., * matches any amount of those

\n matches newline (the end of a line)

In other words: The expression matches any combination of digits or . that ranges from the start of the line to the end of the line

However i am confused why your "Should return" block contains 1.2.1 in the last line, could you explain why 1.2.1 should be matched?

user_115359050

You can match every digit and dot [\d\.]+, which group is found

  • after a boundary or a space (\b|\s)
  • before an end of line character or a space (\s|$)

Here's the complete regex:

(\b|\s)([\d\.]+)(\s|$)

If you want a more strict version, that avoids matching versions that have less than two numbers and more than three, you can instead use the following:

(\b|\s)([\d]+\.[\d]+([\.]+[\d]+)?)(\s|$)

In order to retrieve the version, you can access Group 2 for both regex.

Does it work for your cases?

user_124928900

You can attempt to match the regular expression with the case-indifferent flag set.

^\d+(?:\.\d+)+(?![_\d.]|[a-z][a-z\d]*_)

Demo

The expression contains the following elements.

^           # match beginning of string
\d+         # match >= 1 digits
(?:         # begin non-capture group
  \.\d+     # match a period followed by >= 1 digits
)+          # end non-capture group and execute it >= 1 times
(?!         # begin negative lookahead
  [_\d.]    # match an underscore, digit or period
|           # or
  [a-z]     # match a letter
  [a-z\d]*  # match >= 0 letters or digits
  _         # match underscore
)
user_2569700

You can use

\b\d+(?:\.\d+)*[a-zA-Z]*(?!\S)

Explanation

  • \b A word boundary to prevent a partial word match
  • \d+(?:\.\d+)* Match 1+ digits and optionally repeat matching a dot and 1+ digits
  • [a-zA-Z]* Match optional chars a-zA-Z
  • (?!\S) Assert a whitespace boundary to the right

See a regex demo.

user_54249880
Related