Regex Pattern for phone series 97890x00yz

Question

Regex Pattern for phone series 97890x00yz where x,y,z should not have same values ?

possible duplicate of [Learning Regular Expressions](http://stackoverflow.com/questions/4736/learning-regular-expressions) — Javier, Feb 20 '15 at 10:54
All other things except x,y and z are constant?. i.e, 97890 and 00 are constant or can they change? — TheLostMind, Feb 20 '15 at 10:54

TheLostMind · Answer 1 · 2015-02-20T11:04:26.993

1

Something like this will work for you :

public static void main(String... strings) {
    // String s = "97890x00yz";
       String s = "9789010023"; // output true.
    System.out.println(s.matches("97890(\\d)00(?!\\1)(\\d)(?!\\1|\\2)\\d"));
    //String s = "9789010022"; same regex - output : false
}

edited Feb 20 '15 at 11:04

answered Feb 20 '15 at 10:57

TheLostMind

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can u explain the logic? – gurtej singh Feb 23 '15 at 04:33
`()` captures everything inside it. SO you capture digits and `(?!)` is negative look-ahead.. It makes sure that the mentioned string doesn't follow..`\\1` refers to first captured group (digit) . `\\2` referes to second and so on. – TheLostMind Feb 23 '15 at 06:26
thanx bro...could u help me with this string 978900x00x? – gurtej singh Feb 23 '15 at 09:30
and 9789xyyyyy where x and y having diff values – gurtej singh Feb 23 '15 at 10:17
help me with this regex..9789xyy000...im getting some error – gurtej singh Feb 23 '15 at 10:49

score 0 · Answer 2 · answered Feb 20 '15 at 10:56

0

97890(\d)00(?!\1)(\d)(?!\2|\1)\d

You can try this.See demo.

answered Feb 20 '15 at 10:56

vks

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Regex Pattern for phone series 97890x00yz

2 Answers2