regex Pattern Matcher, how to store the rest of the string

Question

String : USA (45)

using pattern matching getting only numeric value 45

Java code

ArrayList<String> portfolioCount = new ArrayList<String>();
String mainText = USA (45)
final Pattern p = Pattern.compile("\\((.*?)\\)");
final Matcher m = p.matcher(mainText);
m.find();           
portfolioCount.add(m.group(1));

store the numeric value into an arrayList, my question is how to store the rest of the String (ie USA) into another array list

Answer 1

(.*?)\\s*\\((.*?)\\)

Try this.Grab the match1 and match2.

See demo.

http://regex101.com/r/qC9cH4/7

Answer 2

Just capture the part before the number within () to another group,

ArrayList<String> portfolioCount = new ArrayList<String>();
String mainText = "USA (45)";
final Pattern p = Pattern.compile("(\\S+)\\s+\\((.*?)\\)");
final Matcher m = p.matcher(mainText);
while(m.find()){           
portfolioCount.add(m.group(1));
portfolioCount.add(m.group(2));
System.out.println(portfolioCount);
}

Output

[USA, 45]

To store the country name and count into two separate lists,

ArrayList<String> portfolioCount = new ArrayList<String>();
ArrayList<String> country = new ArrayList<String>();
String mainText = "USA (45)";
final Pattern p = Pattern.compile("(\\S+)\\s+\\((.*?)\\)");
final Matcher m = p.matcher(mainText);
while(m.find()){           
portfolioCount.add(m.group(1));
country.add(m.group(2));
System.out.println(portfolioCount);
System.out.println(country);
}

Output:

[USA]
[45]

Answer 3

You can use lookahead and lookbehind in order not to include the parenthesis in the matched group:

ArrayList<String> countryCount = new ArrayList<String>();
ArrayList<String> portfolioCount = new ArrayList<String>();
String mainText = "USA (45)";
final Pattern p = Pattern.compile("(.*?)(?>\\()(\\d+)(?=\\)).*");
final Matcher m = p.matcher(mainText);
m.find();
countryCount.add(m.group(1)); //USA
portfolioCount.add(m.group(2)); //45