Java regex to split a string by using different delimiters

Question

Suppose I want to split a string by either space character or the %20 string, how should I write my regex?

I tried the following, but it didn't work.

String regex = "[\\s+, %20]";
String str1 =  "abc%20xyz";
String str2 = "abc xyz";

str1.split(regex);
str2.split(regex);

The regex doesn't seem to work on str1.

Casimir et Hippolyte · Accepted Answer · 2013-07-10 02:52:35Z

3

use the alternation |:

String regex = "(?:\\s+|%20)+";

answered Jul 10, 2013 at 2:52

Casimir et Hippolyte

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4 Comments

One Two Three Over a year ago

Whats' the meaning of ?:?

Casimir et Hippolyte Over a year ago

@OneTwoThree: (?:...) is for grouping without capturing

One Two Three Over a year ago

Wait, but this is not the same as what I'd want. (I believe this regex means to split by one or more '%20', which is not what I want.)

Casimir et Hippolyte Over a year ago

@OneTwoThree: I am not sure what you are looking for, but you have now all the tools to make your own pattern.

lee tan · Accepted Answer · 2013-07-10 03:12:48Z

1

String regex = "(\\s{1}+|%20{1}+)";

answered Jul 10, 2013 at 3:12

lee tan

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3 Comments

Ro Yo Mi Over a year ago

In your {1}+ is like saying I want exactly one of those characters once or more times?

lee tan Over a year ago

No,{1}+is like say once times,++means once or more times,e.g. "abc%20%20%20xyz".split("([%20]++)").length==2; but "abc%20%20%20xyz".split("(%20+)").length==4;

Ro Yo Mi Over a year ago

Hmm that is interesting. Thanks for piquing my curiosity, I'll have to read up on this.

Alex Guerra · Accepted Answer · 2013-07-10 10:21:15Z

0

If you want to split by ONE space or ONE "%20", try this:

String regex = "(\\s|%20)";

If you want to split by AT LEAST ONE space or AT LEAST ONE "%20", then try this:

String regex = "(\\s+|(%20)+)";

answered Jul 10, 2013 at 10:21

Alex Guerra

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