Grabbing values not between quotation marks (regular expression)

Question

I want to fetch the word outside a quotation for example if I have this text:

  xxxx xxx OK xxxxx "xxx OK xxxx"

I want to fetch the "OK" outside the quotation which mean first one I make this regular expression

  [ ]OK[ ]

but it fetch the both. So how can I fetch only outside a quotation with regular expression?

I will use java, but now I check me regular expiration on this website. — Aymn Alaney, Oct 26 '15 at 02:50
for java use https://regex101.com/r/qJ7eI8/2 and fetch the chars you want from group index 1. — Avinash Raj, Oct 26 '15 at 02:51

score 1 · Accepted Answer · answered Oct 26 '15 at 03:09

You may use the combination of caturing group and the regex alternation operator.

"[^"]*"|( OK )

And now the group index 1 contain the <space>OR<space> which exists outside the double quoted part.

Code:

Matcher m = Pattern.compile("\"[^\"]*\"|( OK )").matcher(s);
while(m.find())
{
System.out.println(m.group(1));
}

1 Answers1