(如果你可以假设
Java> = 9,则4castle的答案比下面的要好)
你需要创建一个匹配器,并使用它来迭代查找匹配项。
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()); }
之后,
allMatches包含匹配项,
allMatches.toArray(new String[0])如果你确实需要一个数组,则可以使用它来获取一个数组。
MatchResult由于Matcher.toMatchResult()返回了当前组状态的快照,因此你还可以编写辅助函数来循环匹配。
例如,你可以编写一个惰性迭代器来完成
for (MatchResult match : allMatches(pattern, input)) { // Use match, and maybe break without doing the work to find all possible matches.}
通过做这样的事情:
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; } public void remove() { throw new UnsupportedOperationException(); } }; } };}
有了这个,
for (MatchResult match : allMatches(Pattern.compile("[abc]"), "abracadabra")) { System.out.println(match.group() + " at " + match.start());}
产量
a at 0b at 1a at 3c at 4a at 5a at 7b at 8a at 10
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