This regex isn't allowing the extra digit for some reason... why?

Question

I'm trying to make it so that the following are able to fit into the regex, but I'm not able to make it work.

These should pass

3
3.4
3.5.3
2.5.*
3.*

These should fail

33.2
2.
3.1.2.
.2
3.1.2.....

My current regex is \d?[\.?]? This only approves 3. I try going forward and use \d?[\.?\d?]? but it still only allows 3.

Edit: 3.5 technically passes in the way that 3. and 5 pass individually but not together. How can I make it together?

Why isn't it working on the rest of the numbers? And how do I fix it?

Update: Please explain how the regex does work if you manage to find it. Thank you.

@evolutionxbox No flags, and my problem is with the regex... not the JS. Is it needed? — 12944qwerty, Sep 18 '20 at 22:24
It’s useful to share it as it may highlight something else you could be overlooking. — evolutionxbox, Sep 18 '20 at 22:25
[Your regex](https://regex101.com/r/b5qBV9/1) matches more than just `3`. — Wiktor Stribiżew, Sep 18 '20 at 22:25
@WiktorStribiżew It does, but I want it to match the full numbers. For example: `3.5` matched, but it was matched like `3.` and `5` — 12944qwerty, Sep 18 '20 at 22:28
@evolutionxbox Sorry, but I actually don't have any javascript code. The error is coming from regex101 directly instead of js. But I believe I'm going to be running the code with `pattern.exec(nums[i])` — 12944qwerty, Sep 18 '20 at 22:29
`(?:\d\.?\*?){1,3}` is my initial attempt. It matches the pattern you want, but I don't know how good it is. — evolutionxbox, Sep 18 '20 at 22:34
Why are you putting `?` inside the brackets? You don't want to match a `?` character. — Barmar, Sep 18 '20 at 22:36
You don't need to escape `.` inside square brackets, it has no special meaning there. — Barmar, Sep 18 '20 at 22:37
@evolutionxbox :O That's close `3.2.3.` passes and I don't want multiple digits between periods (`3.32` passes) — 12944qwerty, Sep 18 '20 at 22:40
@Barmar I'm not very experienced in regex and that's how I thought it would work. the `?` means 0 or 1 instances of the previous character — 12944qwerty, Sep 18 '20 at 22:41
Inside `[]` is not a pattern, it's just a list of characters to match. You put `?` after that to match 0 or 1 of it. — Barmar, Sep 18 '20 at 22:45
`^\d+?(?:\.[\d*]+){0,2}(?!\.$)$` second attempt. https://regex101.com/r/Wcsr2H/1 — evolutionxbox, Sep 18 '20 at 22:51
Wow, this is closest. but `33.2` passes when it shouldn't (only one number in between) — 12944qwerty, Sep 18 '20 at 22:52
For those exact matches `^\d(?:\.\d)?(?:\.[*\d])?$` https://regex101.com/r/gbL2tM/1 — The fourth bird, Sep 18 '20 at 22:57
Has to start with a single digit, may be followed by any number of pairs of period+singledigit, last pair may be period+asterisk: `^\d(\.\d)*(\.\*)?$`. Instead of single digit, any number of digits (at least one): `^\d+(\.\d+)*(\.\*)?$`. — iAmOren, Sep 18 '20 at 22:59
I must say that the question is not very clear, but the pattern `^\d(?:\.\d)?(?:\.[*\d])?$` matches a single digit, then optionally a dot and a single digit, then optionally a dot and an asterix or digit using a character class. To prevent partial matches, it uses an anchor `^` to assert the start of the string, and and anchor `$` to assert the end of the string. — The fourth bird, Sep 18 '20 at 23:05
@Thefourthbird I meant to make an answer so I can accept it. But in that answer explain it lol. But thanks! — 12944qwerty, Sep 18 '20 at 23:08

Matt Davis · Answer 1 · 2020-09-18T22:34:01.353

0

You can use ^(\d+\.)?(\d+\.)?(\*|\d+)$ as found in this answer for version numbering regex.

If you want only a singular digit between the dots then remove the + symbols resulting in:

^(\d\.)?(\d\.)?(\*|\d)$

edited Sep 18 '20 at 22:34

answered Sep 18 '20 at 22:28

Matt Davis

1,167
8
21

This regex code appears to have the same problem as I stated above. (but I want it to match the full numbers. For example: `3.5` matched, but it was matched like `3.` and `5`) – 12944qwerty Sep 18 '20 at 22:34
They're all still part of match 1. Yes `3.` is match 1 capturing group 1 and `5` is match 1 capturing group 2, but the regex results in a single match. – Matt Davis Sep 18 '20 at 22:36
Since he doesn't need to capture the different parts, use non-capturing groups to make it clearer. – Barmar Sep 18 '20 at 22:38
^ `^(?:\d+\.)?(?:\d+\.)?(?:\*|\d+)$` or shortened `^(?:\d+\.){0,2}(?:\*|\d+)$` The single digit version can be further shortened to `^(?:\d\.){0,2}[*\d]$`. – 3limin4t0r Sep 18 '20 at 22:44

iAmOren · Answer 2 · 2020-09-18T23:48:20.570

^\d(\.\d)*(\.\*)?$

var values=[
  "3", "3.4", "3.5.3", "2.5.*", "3.*",
  "33.2", "2.", "3.1.2.", ".2", "3.1.2."
];

values.forEach(value=>
  console.log(
    value,
    value.match(/^\d(\.\d)*(\.\*)?$/)?"Pass":"Fail"
  )
);

function test() {
  var value=document.getElementById("input").value;
  document.getElementById("result").innerText=
    value.match(/^\d(\.\d)*(\.\*)?$/)?"Pass":"Fail";
}

<input id="input" onchange="test()">
<button onclick="test()">Test</button>
<span id="result"></span>

Must start (^) with a single digit (\d).
May be followed by unlimited pairs of period+single digit ((\.\d)*) - pair: (\.\d), unlimited: *.
May be followed by a single pair of period+asterix ((\.\*)?) - pair: (\.\*), may(single/none): ?.
And end: $.

Should you want a non-zero digit, replace all ds with [1-9].

Should you want a number rather than a digit, add a '+' after each d: ^\d+(\.\d+)*(\.\*)?$.
And without leading zeros: ^[1-9]\d*(\.[1-9]\d*)*(\.\*)?$.

dale landry · Answer 3 · 2020-09-19T00:07:04.527

(?:\d\.) -> Non capturing group, with Digit \d matching 0-9. Escaped decimal matches a decimal.

? -> followed by a ? which matches 0 to 1 of the preceding token...

(?:\d\.) -> Non capturing group, with Digit \d matching 0-9. Escaped decimal matches a decimal.

? -> followed by a ? which matches 0 to 1 of the preceding token...

(?:\d) -> Non capturing group, with Digit \d matching 0-9.

$ ends string

NOTE: This will allow 0.*, *.0, 0.0.*, 0.* and 0,0,0 where * = some single digit number... If you wish to filter out zeros as null then more logic would be...

let regEx = /^(?:\d\.)?(?:\*|\d\.)?(?:\*|\d)$/gm;

let el = document.getElementById('input');

el.addEventListener('keyup', function(){
  let thisVal = this.value;
  
  if(thisVal.match(regEx)){
    console.log(thisVal + " is allowed")
  }else(
    console.log(thisVal + " is NOT allowed")
  )
})

Enter a string <input id="input">

3.5.* should be allowed, but isn't with your regex – 12944qwerty Sep 19 '20 at 00:05 — 12944qwerty, Sep 19 '20 at 00:05

score 0 · Answer 4 · answered Sep 28 '20 at 13:10

You could match a digit followed by an optional decimal part and an optional part to match a dot followed by either an asterix or a digit.

To prevent partial matches you can use anchors to assert the start ^ and the $ end of the string.

^\d(?:\.\d)?(?:\.[*\d])?$

Explanation

^ Start of string
\d Match a digit
(?:\.\d)? Optionally match a dot followed by a digit
(?:\.[*\d])? Optionally match a dot and either a * or digit
$ End of string

Regex demo

This regex isn't allowing the extra digit for some reason... why?

4 Answers4