Filter version number from string in javascript?

Question

I found some threads about extracting version number from a string on here but none that does exactly what I want.

How can I filter out the following version numbers from a string with javascript/regex?

Title_v1_1.00.mov filters 1

v.1.0.1-Title.mp3 filters 1.0.1

Title V.3.4A. filters 3.4A

V3.0.4b mix v2 filters 3.0.4b

So look for the first occurrence of: "v" or "v." followed by a digit, followed by digits, letters or dots until either the end of the string or until a whitepace occurs or until a dot (.) occurs with no digit after it.

@Thefourthbird Looking good. How do I make so it also takes the most simple version "v1"? — SeaBass
– SeaBass, Commented Oct 21, 2018 at 8:30
@Thefourthbird No, but a new example would be Title v1 V3.0.4b mix v2 where it should match just 1 for "v1". It seems like you need something after the first digit for it to match, which I realize I may have been misleading in the original post. — SeaBass
– SeaBass, Commented Oct 21, 2018 at 8:33
Then you could use an anchor ^. See this demo Should the [a-z] only be after the last digit like 3.4Aor could it also be 3.4A.5B — The fourth bird
– The fourth bird, Commented Oct 21, 2018 at 8:36

The fourth bird · Accepted Answer · 2018-10-21 15:56:13Z

1

As per the comments, to match the first version number in the string you could use a capturing group:

^.*?v\.?(\d+(?:\.\d+[a-z]?)*)

Regex demo

That will match:

^ Assert the start of the string
.*? Match 0+ any character non greedy
v\.? Match v followed by an optional dot
( Capturing group
- \d+ Match 1+ digits
- (?: Non capturing group
  - \.\d+[a-z]? Match a dot, 1+ digits followed by an optional character a-z
- )* Close non capturing group and repeat 0+ times
) Close capturing group

If the character like A in V.3.4A can only be in the last part, you could use:

^.*?v\.?(\d+(?:\.\d+)*[a-z]?)

const strings = [
  "Title_v1_1.00.mov filters 1",
  "v.1.0.1-Title.mp3 filters 1.0.1",
  "Title V.3.4A. filters 3.4A",
  "V3.0.4b mix v2 filters 3.0.4b"
];

let pattern = /^.*?v\.?(\d+(?:\.\d+[a-z]?)*)/i;
strings.forEach((s) => {
  console.log(s.match(pattern)[1]);
});

edited Oct 21, 2018 at 15:56

answered Oct 21, 2018 at 14:44

The fourth bird

165k16 gold badges61 silver badges75 bronze badges

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

3 Comments

SeaBass Over a year ago

Thank you! My last wish. I need v1a to return 1a and v1ab.1abc to return 1ab.1abc. So a dot always need to be followed by a digit and then optional number of letters after that. Thanks again!!

The fourth bird Over a year ago

Try this demo by adding a character class and repeat that 0+ times

SeaBass Over a year ago

Genius! Thanks so much :D

vrintle · Accepted Answer · 2018-10-21 05:32:06Z

Details:

v - character "v"

(?:\.)? - matches 1 or 0 repetition of "."

Version capturing group

[0-9a-z\.]* - Matches alphanumeric and "." character

[0-9a-z] - ensures that version number don't ends with "."

You can use RegExp.exec() method to extract matches from string one by one.

const regex = /v(?:\.?)([0-9a-z\.]*[0-9a-z]).*/gi;

let str = [
  "Title_v1_1.00.mov filters 1",
  "v.1.0.1-Title.mp3 filters 1.0.1",
  "Title V.3.4A. filters 3.4A",
  "V3.0.4b mix v2 filters 3.0.4b"
];

let versions = [];
let v; // variable to store match

for(let i = 0; i < str.length; i++) {
  // Executes a check on str[i] to get the result of first capturing group i.e., our version number
  if( (v = regex.exec(str[i])) !== null) 
    versions.push(v[1]); // appends the version number to the array
  
  // If not found, then it checks again if there is a match present or not
  else if(str[i].match(regex) !== null) 
    i--; // if match found then it loops over the same string again
}

console.log(versions);

Toto · Accepted Answer · 2018-10-21 10:22:19Z

var test = [
  "Title_v1_1.00.mov filters 1",
  "v.1.0.1-Title.mp3 filters 1.0.1",
  "Title V.3.4A. filters 3.4A",
  "V3.0.4b mix v2 filters 3.0.4b",
];
console.log(test.map(function (a) {
  return a.match(/v\.?([0-9a-z]+(?:\.[0-9a-z]+)*)/i)[1];
}));

Explanation:

/                       # regex delimiter
    v                   # letter v
    \.?                 # optional dot
    (                   # start group 1, it will contain the version number
        [0-9a-z]+       # 1 or more alphanumeric
        (?:             # start non capture group
            \.          # a dot
            [0-9a-z]+   # 1 or more alphanumeric
        )*              # end group, may appear 0 or more times
    )                   # end group 1
/i                      # regex delimiter and flag case insensitive

Collectives™ on Stack Overflow

Filter version number from string in javascript?

3 Answers 3

3 Comments

Comments

Comments

Your Answer

Hot Network Questions

Collectives™ on Stack Overflow

3 Answers 3

3 Comments

Comments

Comments

Your Answer

Sign up or log in

Post as a guest

Related