使用正则表达式在Java中拆分字符串数组

Question

I'm trying to split this string :

aba(2)bb(52)gc(4)d(2)fe(14)f(6)g(8)h(4)5(6)

so it looks like this array :

[ a, b, a(2), b, b(52), g, c(4), d(2), f, e(14), f(6), g(8) ]

Here are the rules, it can accept letters a to g , it can be a letter alone but if there is parentheses following it, it has to include them and its content. The content of the parentheses must be a numeric value .

This is what I tried :

content = "aba(2)bb(52)gc(4)d(2)fe(14)f(6)g(8)h(4)5(6)";
        a = content.split("[a-g]|[a-g]\\([0-9]*\\)");
        for (String s:
             a) {
            System.out.println(s);
        }

And here's the output

(2)

(52)

(4) (2)

(14) (6) (8)h(4)5(6)

Thanks.

Answer 1

It is easier to match these substrings:

String content = "aba(2)bb(52)gc(4)d(2)fe(14)f(6)g(8)h(4)5(6)";
Pattern pattern = Pattern.compile("[a-g](?:\\(\\d+\\))?");
List<String> res = new ArrayList<>();
Matcher matcher = pattern.matcher(content);
while (matcher.find()){
    res.add(matcher.group(0)); 
} 
System.out.println(res);

Output:

[a, b, a(2), b, b(52), g, c(4), d(2), f, e(14), f(6), g(8)]

See the Java demo and a regex demo .

Pattern details

• [ag] - a letter from a to g
• (?:\\(\\d+\\))? - an optional non-capturing group matching 1 or 0 occurrences of
  • \\( - a ( char
  • \\d+ - 1+ digits
  • \\) - a ) char.

Answer 2

If you want to use the split method only, here is an approach you could follow too,

import java.util.Arrays;

public class Test 
{
   public static void main(String[] args)
   {
        String content = "aba(2)bb(52)gc(4)d(2)fe(14)f(6)g(8)h(4)5(6)";
        String[] a = content.replaceAll("[a-g](\\([0-9]*\\))?|[a-g]", "$0:").split(":");
        // $0 is the string which matched the regex

        System.out.println(Arrays.toString(a));

   }

}

Regex : [ag](\\\\([0-9]*\\\\))?|[ag] matches the strings you want to match with (ie a, b, a(5) and so on)

Using this regex I first replace those strings with their appended versions (appended with :). Later, I split the string using the split method.

Output of the above code is,

[a, b, a(2), b, b(52), g, c(4), d(2), f, e(14), f(6), g(8), h(4)5(6)]

NOTE: This approach would only work with a delimiter that is known to not be present in the input string. For example, I chose a colon because I assumed it won't be a part of the input string.

Answer 3

Split is the wrong approach for this, as it is hard to eliminate wrong entries.

Just "match", whatever is valid and process the result array of found matches:

[a-g](?:\(\d+\))?

Debuggex Demo

Answer 4

You can try the following regex: [ag](\\(.*?\\))?

• [ag] : letters from a to g required
• (\\(.*?\\))? : any amout of characters between ( and ) , matching as as few times as possible

You can view the expected output here .

This answer is based upon Pattern , an example:

String input = "aba(2)bb(52)gc(4)d(2)fe(14)f(6)g(8)h(4)5(6)";

Pattern pattern = Pattern.compile("[a-g](?:\\(\\d+\\))?");
Matcher matcher = pattern.matcher(input);
List<String> tokens = new ArrayList<>();
while (matcher.find()) {
    tokens.add(matcher.group());
}

tokens.forEach(System.out::println);

Resulting output:

a
b
a(2)
b
b(52)
g
c(4)
d(2)
f
e(14)
f(6)
g(8)

Edit: Using [ag](?:\\((.*?)\\))? you can also easily extract the inner value of a bracket:

while (matcher.find()) {
    tokens.add(matcher.group());
    tokens.add(matcher.group(1)); // the inner value or null if no () are present 
}