php - Regex to split string if it containt wordA or wordB -

edit again trying make more clear.

wich php regex pattern give me match array containing 2 values wich 2 part of string splitted "worda" or "wordb". if string not containt word, return string first array null in second array.

exemple:

preg_match("pattern","foo worda bar",$match), $match contain array['foo', 'bar'] preg_match("pattern","foo wordb bar",$match), $match contain array['foo', 'bar'] preg_match("pattern","foo bar test",$match), $match contain array['foo bar test', null] 

i know $match first value string don't write it.

old question:

i need split 1 line address part. can't find way capture street part dont include app or apt word if present , if present, capture words after it.

for exemple:

"5847a, rue principal app a" should match: (5847, a, rue principal,a)

"5847a, rue prince arthur apt 22" should match: (5847, a, rue prince arthur, 22)

"1111, sherwood street" should match: (1111, , sherwood street, )

i'm using php.

what have far is: /^(\d+)(.*), (.*)(?:app|apt)(?:\s*(.*))?$/i wich wook exemple 1 , 2. if try make alternative (app|apt) optionnal adding ? after it, third match include word app or apt...

any idea how exclude optionnal , alternative app or apt word match?

thank you

edit:

i can simplify problem: how can regex string match return same string minus word app or apt if present in middle of it.

as @madarauchiha pointed out, it's bad idea run regex on address since can in format.

if know have consistent addresses, guess can use regex:

^([0-9]+)([a-z])?,\s(?:(.*?)\s(?:app|apt)\s(.*)|(.*))$ 

and replace:

$1,$2,$3$5,$4 

here's how it's performing.

it's pretty similar yours (i changed few things) , added or (|) operator address second type of addresses without app or apt.

if want consistent number of matches, maybe this?

^([0-9]*)([a-z]?),((?:(?!\sapp|\sapt).)*)(?:\sapp|\sapt)?(.*)$ 

regex101 example.