How to validate version number using regex pattern

Question

I have to validate version number pattern for following examples:

A1
aabc1
AabC134
aabc12.2
aA1.2.3
0.1.1
0.0.2
a.b.c
a.1.2
a.0.0
1.0.0
1.0
1

Basically it should allow alphanum in all three parts (split parts by dot) but it cannot be:

0
0.0.0
000.000.000
0000.00.00

I have tried this regex but it allows zeros:

/([A-Za-z\d]+)?(.*[A-Za-z\d]+)?(.*[A-Za-z\d]+)$

Can it be modified to achieve above results?

/^(?![0.]+$)[A-Za-z0-9]+(?:\.[A-Za-z0-9]+){0,2}$/, see this demo. — Wiktor Stribiżew
– Wiktor Stribiżew, Commented Jul 15, 2019 at 15:29

Tim Biegeleisen · Accepted Answer · 2019-07-15 15:31:02Z

2

I might just use negative lookaheads to assert that the blacklisted version numbers do not appear, and otherwise proceed along the lines of what you are already doing:

 ^(?!^(?:0|0\.0\.0|000\.000\.000|0000\.00\.00)$)[A-Za-z0-9]+(?:\.[A-Za-z0-9]+){0,2}$

Demo

answered Jul 15, 2019 at 15:31

Tim Biegeleisen

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Wiktor Stribiżew · Accepted Answer · 2019-07-15 15:35:36Z

1

You may use

/^(?![0.]+$)[A-Za-z0-9]+(?:\.[A-Za-z0-9]+){0,2}$/

Or,

/^(?![0.]+$)[A-Z\d]+(?:\.[A-Z\d]+){0,2}$/i

See the regex demo

Details

^ - start of string
(?![0.]+$) - no just zeros / dots till the end of string
[A-Za-z0-9]+ - one or more digits/letters
(?:\.[A-Za-z0-9]+){0,2} - 0, 1 or 2 repetitions of . and 1+ digits or letters
$ - end of string

Regex graph:

edited Jul 15, 2019 at 15:35

answered Jul 15, 2019 at 15:32

Wiktor Stribiżew

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2 Comments

Vicky Gonsalves Over a year ago

That's Perfect for my scenario! Thanks Wiktor Stribiżew

Tim Biegeleisen Over a year ago

While your answer may be streamlined and correct, were this regex to appear inside some application code, I would much rather see a plain sight blacklist of version numbers. Imagine if you move on, and someone else inherits your code. Would that person be able to figure out that (?![0.]+$) were intended to block the four version numbers given in the OP?

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