括号中的字母或数字的正则表达式

Question

I am using Java to process text using regular expressions. I am using the following regular expression

^[\\([0-9a-zA-Z]+\\)\\s]+

to match one or more letters or numbers in parentheses one or more times. For instance, I like to match (aaa) (bb) (11) (AA) (iv) or (111) (aaaa) (i) (V)

I tested this regular expression on http://java-regex-tester.appspot.com/ and it is working. But when I use it in my code, the code does not compile. Here is my code:

    import java.util.regex.Matcher;
    import java.util.regex.Pattern;
    public class Tester {

        public static void main(String[] args) {

            Pattern pattern = Pattern.compile("^[\([0-9a-zA-Z]+\)\s]+");

            String[] words = pattern.split("(a) (1) (c) (xii) (A) (12) (ii)");

            String w = pattern.

            for(String s:words){

                System.out.println(s);

            }
    }
}

I tried to use \\ instead of \\ but the regex gave different results than what I expected (it matches only one group like (aaa) not multiple groups like (aaa) (111) (ii).

Two questions:

1. How can I fix this regex and be able to match multiple groups?
2. How can I get the individual matches separately (like (aaa) alone and then (111) and so on). I tried pattern.split but did not work for me.

Answer 1

Firstly, you want to escape any backslashes in the quotation marks with another backslash. The Regex will treat it as a single backslash. (Eg call a word character \\w in quotation marks, etc.)

Secondly, you got to finish the line that reads:

String w = pattern.

That line explains why it doesn't compile.

Answer 2

Here is my final solution to match the individual groups of letters/numbers in brackets that appear at the beginning of a line and ignore the rest

import java.util.ArrayList;
import java.util.regex.Matcher;
import java.util.regex.Pattern;

public class Tester {
    static ArrayList<String> listOfEnums;
    public static void main(String[] args) {
        listOfEnums = new ArrayList<String>();
        Pattern pattern = Pattern.compile("^\\([0-9a-zA-Z^]+\\)");
        String p = "(a) (1) (c) (xii) (A) (12) (ii) and the good news (1)";
        Matcher matcher = pattern.matcher(p);
        boolean isMatch = matcher.find();
        int index = 0;
        //once you find a match, remove it and store it in the arrayList. 
        while (isMatch) {
          String s = matcher.group();
          System.out.println(s);
          //Store it in an array
          listOfEnums.add(s);
          //Remove it from the beginning of the string.
          p = p.substring(listOfEnums.get(index).length(), p.length()).trim();
          matcher = pattern.matcher(p);
          isMatch = matcher.find();
          index++;
        }
    }
}

Answer 3

1) Your regex is incorrect. You want to match individual groups of letters / numbers in brackets, and the current regex will match only a single string of one or more such groups. Ie it will match

(abc) (def) (123)

as a single group rather than three separate groups.

A better regex that would match only up to the closing bracket would be

\([0-9a-zA-Z^\)]+\)

2) Java requires you to escape all backslashes with another backslash

3) The split() method will not do what you want. It will find all matches in your string then throw them away and return an array of what is left over. You want to use matcher() instead

Pattern pattern = Pattern.compile("\\([0-9a-zA-Z^\\)]+\\)");
Matcher matcher = pattern.matcher("(a) (1) (c) (xii) (A) (12) (ii)");

while (matcher.find()) {
  System.out.println(matcher.group());
}