Regex to find evaluate number pattern

Question

I'm trying to create a regular expression to match on a specific given number pattern.

The patterns are:

123456 1234XX 1234*

Not allowed are: A combinations like: 1234*X or 1234X* Or a multiple *: 1234**

I already tried this expression:

/^[\\+\\#\\*0-9]{0,25}[X]*$/

But here I got the multiple * and 1234*X as valid expressions.

Does anyone have an idea of a proper soultion or a way to a solution?

Edit: Ok, I think I should clearify the rules.

The regex should match any given number:

012345
+12345
#1234525

But also match for strings with a X in it.

12345X
12345XX
1234XXXXXX
XXXXXX

The Xs should always stand on the end of the string.

A single * should only be allowed at the end of the string.

1234*
1234556*
1*

Multiple * aren't allowed. Only one * at the end of the string.

The combination of X and * are not allowed. 1234X* or 12345*X are not allowed.

Length restriction: 0 to 25 characters.

Try `'~^(?=.{0,25}$)\d+[X*]*$~'`, or [`'~^(?=.{0,25}$)\d+[X*]{2}$~'`](https://regex101.com/r/dZ0sP6/1), but I don't get the specs from these examples. Please **clarify** the rules. — Wiktor Stribiżew, Aug 31 '16 at 09:53
Try [`'~^(?=.{0,25}$)[+#]?\d*(?:X*|\*?)$~'`](https://regex101.com/r/dZ0sP6/2) - this will make sure the whole string is 0 to 25 chars long, and will only allow one `*` at the end of the string, and zero or more `X` can only follow 0 or more digits. **Please check**. — Wiktor Stribiżew, Aug 31 '16 at 11:19
It looks like this `^[0-9]{0,25}[*]{0,1}$|^[0-9]{0,25}[X]*$` would resolve my problem need to test it though. — legionth, Aug 31 '16 at 11:19

Toto · Answer 1 · 2016-08-31T10:26:44.800

1

As far as I understand your needs, this would do the job:

^\d+(?:x+|\*)?$

New version according to question edit:

^[+#]?\d*(?:x+|\*)?$

edited Aug 31 '16 at 10:26

answered Aug 31 '16 at 10:16

Toto

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revo · Answer 2 · 2016-08-31T10:30:25.810

Based on your new update:

^(?=.)[+#*]?\d*?(?:X+|\*)?$

Live demo

Explanation:

^           # Asserts beginnig of input string
(?=.)       # Positive looakehad - input string length should be at least 1
[+#*]?      # Optional symbols 
\d*?        # Optional digits (any number)
    (?:         # Start of non-capturing group (a)
        X+          # Any number of X
        |           # Or
        \*          # A single *
    )?          # End of non-capturing group (a) - optional
$           # End of string

score 0 · Accepted Answer · answered Aug 31 '16 at 11:24

You can use

'~^(?=.{0,25}$)[+#]?\d*(?:X*|\*?)$~D'

See regex demo

This will make sure the whole string is 0 to 25 chars long, and will only allow one * at the end of the string, and zero or more X can only follow 0 or more digits.

Details:

^ - start of string
(?=.{0,25}$) - a positive lookahead checking the whole string length (must be from 0 to 25 chars long)
[+#]? - one or zero + or #
\d* - zero or more digits
(?:X*|\*?) - a non-capturing group matching either of the 2 alternatives:
- X* - zero or more X symbols
- | - or...
- \*? - one or zero * symbols
$ - end of string (better replace it with \z to ensure no newline at the end is accepted, or just add D modifier).

Please note I used the `D` modifier in the regex, it will make sure the `$` matches the very end of the string. Without it, `$` may match before the final newline symbol which might not be good when validation is performed. Note you can also replace `$` with `\z` anchor to achieve the same behavior. — Wiktor Stribiżew, Aug 31 '16 at 11:28

Regex to find evaluate number pattern

3 Answers3