Java regex split string after number

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+)");

Answer 1

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

Answer 2

You can use this regex fpr splitting:

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

Working Demo

Answer 3

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 (§).