Matching a "." in java

Question

I have the below string

String srcString = "String1.String2.String3";

I want to split "srcString" on "."

Using srcString.split(".") is matching all the characters.

What is the regex to match a "." ?

`String.split("\\.")` It's a regex, not a string literal. – Marko Topolnik Jul 09 '12 at 14:16 — Marko Topolnik, Jul 09 '12 at 14:16

Pshemo · Accepted Answer · 2020-08-26T18:30:57.930

10

In regex dot is special character representing any character except line separator (to also make it match line separators use Pattern.DOTALL flag).

Anyway, use split("\\.")

Explanation:

to escape . we can add \ before it so we end up with regex \.
now since \ is also special in string literal " " we also need to escape it there, so to express \. we need to write it as "\\.".

edited Aug 26 '20 at 18:30

answered Jul 09 '12 at 14:17

Pshemo

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score 3 · Answer 2 · answered Jul 09 '12 at 14:18

3

Use split("\\.") as . (dot) is special character so use \\ before .(dot)

answered Jul 09 '12 at 14:18

Pramod Kumar

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score 0 · Answer 3 · answered Jul 09 '12 at 14:30

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You could also call the function org.apache.commons.lang.StringUtils.split(String, char) of the library commons-lang [http://commons.apache.org/lang/][1]

answered Jul 09 '12 at 14:30

M. Abbas

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Matching a "." in java

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