RegEx ignore letters until digit occurs and then match letters afterwards

Question

This is intended to be used in Java. Imagine following sample input:

I want to match all alphabetic characters after at least one digit.

Desired output:

S
CV

Actual output:

0S
3CV

My current approach looks like this:

\d[a-zA-Z]+

The problem with this pattern is that it includes the digit beforehand too. My current solution is to remove the first character of the resulting string afterwards. And this seems quite unsatisfactory to me.

I think, you'll need groups. This answer should solve you problem: https://stackoverflow.com/a/17969620/4717152 — Andreas M. Oberheim, May 29 '19 at 11:43

score 3 · Accepted Answer · answered May 29 '19 at 11:43

3

You need a lookbehind:

(?<=\d)[a-zA-Z]+

(?<=\d) means "there must be a digit before this position, but don't match it".

Demo

Alternatively, you can use a pair of () to surround the part you want to get:

\d([a-zA-Z]+)

This is called a "group", and you can get its value by calling group(1) on your Matcher.

answered May 29 '19 at 11:43

Sweeper

213,210
22
193
313

Just a side note: `\d([a-zA-Z]+)` needs another backslash in the beginning when using with java. Otherwise it's an invalid escape sequence. Works perfectly fine though! – bautista May 29 '19 at 12:19

score 2 · Answer 2 · answered May 29 '19 at 11:44

2

If you 'add' groups you can get group 1 that contain only letters

\d([a-zA-Z]+)

answered May 29 '19 at 11:44

Egeo

140
7

RegEx ignore letters until digit occurs and then match letters afterwards

2 Answers2

Demo