CODEWITHSUNDEEP
javaregex
Paulo
Paulo

Reputation: 3008

Regex is returning undesired results

I am trying to capture a group from regex that will follow this pattern:

Ex1 - anyanyany group 1 have to be anyanyany

Ex2 - anyanyany.abcany group 1 have to be anyanyany

Ex3 - anyany.abcde.fghi group 1 have to be anyany.abcde

When I try (.+)(?:\.), it only returns Ex2 and Ex3. If I change it for (.+)(?:\.)* it returns the same string of input.

I really don't know what I have to do to solve it. Someone could help me? Which knowledgement I am missing?

https://regex101.com/r/jG6wY8/2

Upvotes: 0

Views: 61

Answers (4)

Eder
Eder

Reputation: 1884

Try the following regex patterns in order to match your criterias:

  • If you were supposed to match the first two words tokenized by a dot: ^([^\.]+)(?:\.[^\.]+)?$|(?:([^\.]+\.[^\.]+)\.)

    a => a a.b => a a.b.c => a.b a.b.c.d => a.b

  • If you were supposed to match every words tokenized by a dot, but not the last token: ^([^\.]+)(?:\.[^\.]+)?$|(?:([^\.]+\.[^\.]+)\.)

    a => a a.b => a a.b.c => a.b a.b.c.d => a.b.c

Upvotes: 0

anubhava
anubhava

Reputation: 784958

You can use this regex:

^([^.]+(?:\.[^.]+)?)

RegEx Demo

PS: Used ^([^.\n]+(?:\.[^.\n]+)?) in regex101 demo since demo has multiple inputs in different lines.

Upvotes: 3

laune
laune

Reputation: 31290

Rather than a regex to capture the wanted part, eliminate the unwanted part:

s = s.replaceAll("\\.[^.]+$","");

Upvotes: 0

Avinash Raj
Avinash Raj

Reputation: 174696

Try this non-greedy regex.

(.+?)(?:\.[^.]*)?$

In java you need to escape the backslash one more time, so it would be like,

Pattern p = Pattern.compile("(.+?)(?:\\.[^.]*)?$");

DEMO

Upvotes: 2

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