正则表达式按第一个数字分割

Question

Note: This is NOT a duplicate of Regex to get first number in string: 100 2011-10-20 14:28:55 (only works when the string starts with a number).

Say I have the following input. The distribution of letters and numbers can be completely random. Additionally, it may contain other characters such as - / ? etc..

'ONEac123TWO45THREEabc67FOUR89bcFIVE'

What I need is an array in the form:

[everything before the first number, the first number, everything else]

So for my example, that would be:

['ONEac','123','TWO45THREEabc67FOUR89bcFIVE']

Thanks in advance :)

Answer 1

You could use .match to get your result.

> 'ONEac123TWO45THREEabc67FOUR89bcFIVE'.match(/^(\D*)(\d+)(.*)$/).slice(1)
  ["ONEac", "123", "TWO45THREEabc67FOUR89bcFIVE"]

Answer 2

How about:

^(\D+)(\d+)(.+)$

http://rubular.com/r/kMhyKH3Sbd

^            start of string
(\D+)        match anything that is not a digit into one group (1 or more, use * for 0 or more)
(\d+)        match 1 or more digits into another group (I am assuming you want ints)
(.+)         match whatever is left into another group
$            end of string