regex matching conditional strings

Question

For example, if I have the following strings:

99%89 (should match)
99%? (should match)
?%99 (should match)
?%? (should not match)
?%99%99 (should match)
99%99%99%? (should match)

essentially the first or second element can be a ? or a number, but both elements cannot be ?. I tried thinking of something like:

[0-9]*|[?](?!\?)[%][0-9]*|[?]

But this does not yield the correct answer, any help would be appreciated

The ? in the negative look ahead needs to be escaped... or I think. Not really clear what the question is. There is no & in the sample strings. — user3783243
– user3783243, Commented Apr 11, 2021 at 1:24
sorry, I edited the question to make it more understandable. even after escaping the ?, it still matches ?%? — AdamantiumPrime
– AdamantiumPrime, Commented Apr 11, 2021 at 1:34
And it shouldn't? Please define what should and shouldn't match, and what it currently does. Question is still not clear to me. Also where regex will be used may be useful. PCRE can do things POSIX can't etc. — user3783243
– user3783243, Commented Apr 11, 2021 at 1:36

RavinderSingh13 · Accepted Answer · 2021-04-11 05:20:42Z

5

With your shown samples, could you please try following.

^(?:(?:\?(?:(?:%\d+){1,})?)|(?:(?:(?:\d+%){1,})?\?(?:(?:%\d+){1,})?)|(?:\d+%\d+))$

Online demo for above regex

Explanation: Adding detailed explanation for above.

^(?:                       ##Matching from starting of the value, starting a non-capturing group from here.
 (?:\?                     ##Starting non-capturing group(one for understanding purposes) matching literal ? here.
    (?:(?:%\d+){1,})?      ##In a non capturing group looking for % with 1 or more occurrences of digits and matching this group match keeping it optional.
  )|                       ##Closing one non-capturing group here, with OR condition here.
 (?:                       ##Starting non-capturing group(two) here.
    (?:(?:\d+%){1,})?\?    ##Looking for digits with % one or more occurrences in a non-capturing group keeping it optional followed by ?
    (?:(?:%\d+){1,})?      ##Checking for % digits one or more occurrences in a non-capturing group keeping it optional followed by ?
 )|                        ##Closing two non-capturing group here, with OR condition here.
 (?:\d+%\d+)               ##In a non-capturing group looking for 1 or more digits % one or more digits
)$                         ##Closing  1st non-capturing group at the end of value.

edited Apr 11, 2021 at 5:20

answered Apr 11, 2021 at 3:23

RavinderSingh13

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3 Comments

RavinderSingh13 Over a year ago

@AdamantiumPrime, you mean more than 1 instances(continuous ones)? Could you please give me a single sample which you tried on? Because with your shown samples my code has worked perfectly fine, let me know.

RavinderSingh13 Over a year ago

@AdamantiumPrime, How about using ^(?:(?:\?(?:(?:%\d+){1,})?)|(?:(?:(?:\d+%){1,})?\?(?:(?:%\d+){1,})?)|(?:\d+%\d+))$ once? I tested it with samples like: 9%?%100%100 99%89 99%? ?%99 ?%? ?%99%99 99%99%99%? Let me know how it goes? See its regex DEMO once regex101.com/r/lLkGe4/1 and let me know then.

RavinderSingh13 Over a year ago

@AdamantiumPrime, Your welcome, could you please update your samples(since you have requested to edit regex for few other cases too, not sure why that comment got deleted) in your question so that it will be clear for people, for understanding purposes, I have updated my answer and explanation of it as well now.

The fourth bird · Accepted Answer · 2021-04-11 17:14:48Z

3

Not sure if I am reading the question right, but as you tried using a negative lookahead you could assert that the string does not only contains % and/or ?

^(?![%?]+$)[\d?%]+$

Regex demo

Or without a lookahead:

^[%?]*\d[%?\d]*$

Regex demo

edited Apr 11, 2021 at 17:14

answered Apr 11, 2021 at 14:33

The fourth bird

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