Java regex split string after number [duplicate]

Question

Sorry, yet another regex question! I have a string of unknown length that contains the following pattern:

"abc1defg2hijk23lmn19"

I need to split the string to get the following:

["abc1", "DefG2", "hiJk23", "lmn19"]

Its pretty simple I just need to split after the number each time. The number can be any non zero number.

I can do a positive lookup like this:

a.split("(?=\\d+)");

However, that actually splits out the number rather than the group of letter in front and the number.

My current attempt looks like this but doesn't work:

a.split("(?=[A-Za-z]+\\d+)");

Pshemo · Accepted Answer · 2014-06-26 14:06:51Z

So for input

abc1defg2hijk23lmn19

you want to find places marked with |

abc1|defg2|hijk23|lmn19

In other words you want to find place which

has digit before (?<=[0-9])
and alphabetic after (?=[a-zA-Z]) it

(I assume you are familiar with look-around mechanisms).

In that case use

split("(?<=[0-9])(?=[a-zA-Z])")

which means place between digit and alphabetic character like 1|a where | represents such place.

Example:

String data ="abc1defg2hijk23lmn19"; for (String s: data.split("(?<=[0-9])(?=[a-zA-Z])")) System.out.println(s);

Output:

abc1 defg2 hijk23 lmn19

anubhava · Accepted Answer · 2014-06-26 13:51:01Z

3

You can use this regex fpr splitting:

(?<=\\d)(?=\\D)

Working Demo

answered Jun 26, 2014 at 13:51

anubhava

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Comments

Adam Yost · Accepted Answer · 2014-06-26 14:08:15Z

replace the number with the number + a space, then split on space.

a = a.replaceAll("[0-9]+","$1 "); String[] b = a.split(" ");

It can also be customized in case your string contains other spaces, by substituting a character guaranteed not to appear in your string instead of the space. my favorite is (§).

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