Get all not matching characters from a regex

Question

I have a regex /[A-Za-z./-]+, and the text that I want to test is ABCD123. After the text doesn't pass the regex, I want to display the characters that are not contained at all in the regex. So in this case, I want to display 123 or (1,2,3). How can I do this ?

The question is for java.

LE: The code accepts general regexes and general texts so it has to be a general solution.

What language are you using? Rather just match everything that isn't a letter. Then you have your 123 (or 1,2,3) — Daniel Lee, Aug 19 '16 at 08:14
The code accepts general regexes and general texts so it has to be a general solution. — Visan Cosmin, Aug 19 '16 at 08:19
There are various flavours of regex, `([^a-zA-Z]{1,})` will work on bigquery. — Daniel Lee, Aug 19 '16 at 08:23
Possible duplicate of [How to extract a substring using regex](http://stackoverflow.com/questions/4662215/how-to-extract-a-substring-using-regex) — xenteros, Aug 19 '16 at 08:31

score 2 · Answer 1 · answered Aug 19 '16 at 08:21

2

You can also first check that regexp pass, when not - remove characters by another regexp, and then you will be have only that not passed characters.

Pseudo code:

if string is not '[A-Za-z./-]+' then
    wrong_chars = string.remove('[A-Za-z./-]+')

answered Aug 19 '16 at 08:21

Marek Skiba

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Can you remove from a String characters specified by a regex ? Because if this is the case, then this is the solution. – Visan Cosmin Aug 19 '16 at 08:23
First result from Google: http://stackoverflow.com/questions/11796985/java-regular-expression-to-remove-all-non-alphanumeric-characters-except-spaces – Marek Skiba Aug 19 '16 at 08:25
The regex can start for example with a "/", and then allow for letter. How do I eliminate only the letters ? The regex doesn't have any method "regex.contains("A")", so how can I elminiate A from ABC if the regex is "/[A]" ? – Visan Cosmin Aug 19 '16 at 08:43

Daniel Lee · Answer 2 · 2020-03-24T14:38:15.183

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([^a-zA-Z]{1,})

Will match your pattern.

As will, more succinctly:

 [^a-zA-Z]+

edited Mar 24 '20 at 14:38

answered Aug 19 '16 at 08:24

Daniel Lee

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Why does this work and `([^a-zA-Z]*)` does not? – Steve Van Opstal Mar 24 '20 at 14:09
* matches beween 0 and unlimited times while + matches between 1 and unlimited. Obvously, we don't want the match of 0 in this case. – Daniel Lee Mar 24 '20 at 14:39
It took me a while to see but negatively matching 0 times indeed doesn't make sense. Thanks – Steve Van Opstal Mar 24 '20 at 19:37

Shekhar Khairnar · Answer 3 · 2016-08-19T08:45:17.917

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You need to just negate your regex :/([A-Za-z.\/-]+)|([^A-Za-z.\/-]+)/g

Second capturing group is the group which will have elements not matched.

Example

edited Aug 19 '16 at 08:45

answered Aug 19 '16 at 08:16

Shekhar Khairnar

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FYI right side of alternation can become shorter to `(.)` – revo Aug 19 '16 at 08:47
You didn't include the first /. Why not ? Beside, the expression can be general and I receive it as a parameter. – Visan Cosmin Aug 19 '16 at 10:03
If your regex has a must on using slash at the very beginning then your input string `ABCD123` is never matched and expected result should be A,B,C,D,1,2,3 and not only 1,2,3. You should clarify this problem. @VisanCosmin – revo Aug 19 '16 at 11:30

revo · Answer 4 · 2016-08-19T08:21:53.283

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Include an alternation which captures all other characters:

[A-Za-z.\/-]+|(.)

This way if left side of alternation does not match whole input string then first capturing group has some values in it.

Live demo

edited Aug 19 '16 at 08:21

answered Aug 19 '16 at 08:16

revo

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Get all not matching characters from a regex

4 Answers4