Create array of regex matches
(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
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()
)
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())