(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())
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;
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.
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"]
Source: Stackoverflow.com