Split and Maintain Delimiter

Question

September 1997, such and such

I need to split the above text into

September 1997
such and such

The delimiter is 1997, that is

/[0-9][0-9][0-9][0-9]\,/

How can I do it in ruby via regex?

That's my attempt:

"September 1997, such and such".to_s.split(/\s+(?=[0-9][0-9][0-9][0-9]\,)/)

Output:

September
1997, such and such

What I want is:

September 1997
such and such

Answer 1

Splitting September 1997, such and such like this

(?<=\d{4}),\s*

should get this output

September 1997
such and such 

You could even add extra whitespace

(?<=\d{4})\s*,\s*

for variability and trimming.

Answer 2

Note this only splits once, but that seems to be what you want?

str = "September 1997, such and such"
p str.match(/(.*?)([0-9]{4}.*)/)
# => #<MatchData "September 1997, such and such" 1:"September " 2:"1997, such and such">

This does do multiple splits (edited to include \b thanks to p11y)

p str.split(/(?=\b\d{4}\b)/)
# => ["September ", "1997, such and such"]

Answer 3

Using positive lookahead, you can split at the empty string, followed by a decimal (\d). In this case that means that the empty string must be followed by a decimal, but the decimal will not be part of the split.

"September 1997, such and such".split(/(?=\d)/)
#=> ["September ", "1997, such and such"]

If you need to match exactly four decimals you can use a quantifier ({n}), e.g.

"September 1997, such and such".split(/(?=\d{4})/)
#=> ["September ", "1997, such and such"]

If you want to get rid of the space in between, you can additionally capture one or more whitespace characters (\s+) instead of the empty string:

"September 1997, such and such".split(/\s+(?=\d)/)
#=> ["September", "1997, such and such"]