'Create array of regex matches
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 form an array of all string matching a regex expression in a given string?
Solution 1:[1]
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())
Solution 2:[2]
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"]
Solution 3:[3]
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())
Solution 4:[4]
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.
Solution 5:[5]
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;
Sources
This article follows the attribution requirements of Stack Overflow and is licensed under CC BY-SA 3.0.
Source: Stack Overflow
| Solution | Source |
|---|---|
| Solution 1 | 4castle |
| Solution 2 | zb226 |
| Solution 3 | walkeros |
| Solution 4 | Anthony Accioly |
| Solution 5 | Nikhil Kumar K |
