Why does split(".") fail? java

Question

I've written the following code:

String[] arr = ((String) "asd.asd").split(".");

and arr=[]. Why?

Because the split function expects a regular expression – Joel Sep 23 '14 at 14:05 — Joel, Sep 23 '14 at 14:05

score 5 · Accepted Answer · edited Sep 23 '14 at 14:08

5

split takes a regular expression as an argument. "." in regular means "any character".

Instead, use:

 String[] arr = "asd.asd".split("\\.");

The backslashes escape the special meaning of the "." character in a regular expression.

http://docs.oracle.com/javase/tutorial/essential/regex/

edited Sep 23 '14 at 14:08

ajp15243

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answered Sep 23 '14 at 14:04

khelwood

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score 2 · Answer 2 · answered Sep 23 '14 at 14:04

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split() accepts a regex. you should escape the . use "\\." . In regex . is a special character (Meta character) which means match any character.

answered Sep 23 '14 at 14:04

TheLostMind

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score 1 · Answer 3 · answered Sep 23 '14 at 14:04

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You must double escape the ., otherwise the regular expression represents it as "any character".

Also, you don't need to cast "asd.asd" as String.

String[] arr = "asd.asd".split("\\.");

answered Sep 23 '14 at 14:04

Mena

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score 1 · Answer 4 · answered Sep 23 '14 at 14:04

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Because '.' is a special character. You need to escape it by writing it like this '\\.'

answered Sep 23 '14 at 14:04

Le Ish Man

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No, forward slashes really aren't going to help. And "special character" really doesn't describe what's going on. – Jon Skeet Sep 23 '14 at 14:05

Why does split(".") fail? java

4 Answers4