Java - regex parse string

Question

Trying to parse out names with given samples

++++++++++++++++++SELIZABETH+COLLAZO+++++++++++++++++++
+++++++++++++++++++PALOMA+CORREA+++++++++++++++++++++++
+++++++++++++++++++NOAH+BLAKEMORE++++++++++++++++++++++

I've tried

//++(.*?)+(.*?)//++

but that's way off.

Would like to parse out the first and last name to two strings.

Did you mean to escape the `+` with `\\ ` instead of using `//`, and also to escape the one in the middle? — Bernhard Barker, Nov 07 '17 at 15:07
To match a literal `+`, you should escape it with ``\\`` in the Java string literal — Wiktor Stribiżew, Nov 07 '17 at 15:09
`+` has special meaning in a regex, you have to escape them: `\\++(.*?)\\+(.*?)\\++` — Toto, Nov 07 '17 at 15:09
You can use `^\++((?:[^+\s]*\+)+[^+\s]+)+` assuming an unknown number of names are possible and then split the resulting group on `+` — ctwheels, Nov 07 '17 at 15:15

score 5 · Answer 1 · answered Nov 07 '17 at 15:09

You can use this regex (\w+)\+(\w+) or \+{2,}(.*?)\+(.*?)\+{2,} with Pattern like this :

String str = "++++++++++++++++++SELIZABETH+COLLAZO+++++++++++++++++++\n"
        + "+++++++++++++++++++PALOMA+CORREA+++++++++++++++++++++++\n"
        + "+++++++++++++++++++NOAH+BLAKEMORE++++++++++++++++++++++";

Pattern pattern = Pattern.compile("(\\w+)\\+(\\w+)");// or instead "\\+{2,}(.*?)\\+"(.*?)\\+{2,}
Matcher matcher = pattern.matcher(str);

while (matcher.find()) {
    System.out.println(matcher.group(1) + " " + matcher.group(2));
}

Outputs

SELIZABETH COLLAZO
PALOMA CORREA
NOAH BLAKEMORE

Hi down-voter, please put a comment when you down-vote a post, it is not legal to down-vote just like this, you have to respect the efforts of others, because we respect your efforts also. Thank you. — Youcef LAIDANI, Feb 19 '20 at 11:25

Java - regex parse string

1 Answers1