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Create array of regex matches

开发者 https://www.devze.com 2023-03-06 01:49 出处:网络
In Java, I am trying to return all regex matches to an array but it seems that you can only check whether the pattern matches something or not (boolean).

In Java, I am trying to return all regex matches to an array but it seems that you can only check whether the pattern matches something or not (boolean).

How can I use a regex match to 开发者_如何学Goform an array of all string matching a regex expression in a given string?


(4castle's answer is better than the below if you can assume Java >= 9)

You need to create a matcher and use that to iteratively find matches.

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

 ...

 List<String> allMatches = new ArrayList<String>();
 Matcher m = Pattern.compile("your regular expression here")
     .matcher(yourStringHere);
 while (m.find()) {
   allMatches.add(m.group());
 }

After this, allMatches contains the matches, and you can use allMatches.toArray(new String[0]) to get an array if you really need one.


You can also use MatchResult to write helper functions to loop over matches since Matcher.toMatchResult() returns a snapshot of the current group state.

For example you can write a lazy iterator to let you do

for (MatchResult match : allMatches(pattern, input)) {
  // Use match, and maybe break without doing the work to find all possible matches.
}

by doing something like this:

public static Iterable<MatchResult> allMatches(
      final Pattern p, final CharSequence input) {
  return new Iterable<MatchResult>() {
    public Iterator<MatchResult> iterator() {
      return new Iterator<MatchResult>() {
        // Use a matcher internally.
        final Matcher matcher = p.matcher(input);
        // Keep a match around that supports any interleaving of hasNext/next calls.
        MatchResult pending;

        public boolean hasNext() {
          // Lazily fill pending, and avoid calling find() multiple times if the
          // clients call hasNext() repeatedly before sampling via next().
          if (pending == null && matcher.find()) {
            pending = matcher.toMatchResult();
          }
          return pending != null;
        }

        public MatchResult next() {
          // Fill pending if necessary (as when clients call next() without
          // checking hasNext()), throw if not possible.
          if (!hasNext()) { throw new NoSuchElementException(); }
          // Consume pending so next call to hasNext() does a find().
          MatchResult next = pending;
          pending = null;
          return next;
        }

        /** Required to satisfy the interface, but unsupported. */
        public void remove() { throw new UnsupportedOperationException(); }
      };
    }
  };
}

With this,

for (MatchResult match : allMatches(Pattern.compile("[abc]"), "abracadabra")) {
  System.out.println(match.group() + " at " + match.start());
}

yields

a at 0
b at 1
a at 3
c at 4
a at 5
a at 7
b at 8
a at 10


In Java 9, you can now use Matcher#results() to get a Stream<MatchResult> which you can use to get a list/array of matches.

import java.util.regex.Pattern;
import java.util.regex.MatchResult;
String[] matches = Pattern.compile("your regex here")
                          .matcher("string to search from here")
                          .results()
                          .map(MatchResult::group)
                          .toArray(String[]::new);
                    // or .collect(Collectors.toList())


Java makes regex too complicated and it does not follow the perl-style. Take a look at MentaRegex to see how you can accomplish that in a single line of Java code:

String[] matches = match("aa11bb22", "/(\\d+)/g" ); // => ["11", "22"]


Here's a simple example:

Pattern pattern = Pattern.compile(regexPattern);
List<String> list = new ArrayList<String>();
Matcher m = pattern.matcher(input);
while (m.find()) {
    list.add(m.group());
}

(if you have more capturing groups, you can refer to them by their index as an argument of the group method. If you need an array, then use list.toArray())


From the Official Regex Java Trails:

        Pattern pattern = 
        Pattern.compile(console.readLine("%nEnter your regex: "));

        Matcher matcher = 
        pattern.matcher(console.readLine("Enter input string to search: "));

        boolean found = false;
        while (matcher.find()) {
            console.format("I found the text \"%s\" starting at " +
               "index %d and ending at index %d.%n",
                matcher.group(), matcher.start(), matcher.end());
            found = true;
        }

Use find and insert the resulting group at your array / List / whatever.


        Set<String> keyList = new HashSet();
        Pattern regex = Pattern.compile("#\\{(.*?)\\}");
        Matcher matcher = regex.matcher("Content goes here");
        while(matcher.find()) {
            keyList.add(matcher.group(1)); 
        }
        return keyList;
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