Regex in java matching digits, few special chars and alphabets

Question

I have a phone number which can contain below values:

• Digits: 0-9
• Special character: "#", "*", "-", "_", " ", "(", ")", "+", "?"
• Alphabets: "x", "X"

These above can be more than once in phone number. If any other value/character besides listed in above list is present it should fail.I tried creating regex but it fails with message "Dangling meta character '*'".

Below is the regex tried:

 if(!str.matches("([0-9]*|*|#|(|)|?|+|_|-)"))
   {
       System.out.println("--not matched--");
   }

Please help me in creating a regex for above.

It's about escaping special chars but I need to match digits, few special chars and only two alphabets. Please once go through my question and help me. Thanks in advance! — Snehal Gupta, May 27 '18 at 06:58
Does the regex `([0-9]*|\*|#|\(|\)|\?|\+|_|-)` do what you want? If no, why not? — Aran-Fey, May 27 '18 at 06:59

score 1 · Answer 1 · edited May 27 '18 at 09:22

1

You need to escape the characters *,+, ? as they have regex meaning.

You also need to escape the \ because you use java, so try

str.matches("([0-9]*|\\*|#|\\(|\\)|\\?|\\+|_|-)")

edited May 27 '18 at 09:22

Aran-Fey

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answered May 27 '18 at 07:13

Yossi Vainshtein

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While this compiles, I'm pretty sure it will not match what OP want. Notice that `matches` expect from regex to match entire string. – Pshemo May 27 '18 at 07:14
@Pshemo-[0-9*#()?+_\\- ]* answer given by you matches my requirement. Thank you!! – Snehal Gupta May 27 '18 at 07:24

Regex in java matching digits, few special chars and alphabets

1 Answers1