创建正则表达式匹配数组

在Java中，我试图将所有正则表达式匹配返回到一个数组，但似乎您只能检查模式是否匹配某些内容（布尔值）。

如何使用正则表达式匹配来形成与给定字符串中的正则表达式匹配的所有字符串的数组？

(4城堡的回答 https://stackoverflow.com/a/46859130如果您可以假设 Java >= 9，则比下面的更好）

您需要创建一个匹配器并使用它来迭代地查找匹配项。

 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());
 }

在这之后，allMatches包含匹配项，您可以使用allMatches.toArray(new String[0])如果您确实需要一个数组，则获取一个数组。

您还可以使用MatchResult编写辅助函数来循环匹配 自从Matcher.toMatchResult()返回当前组状态的快照。

例如，你可以编写一个惰性迭代器来让你做

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

通过这样做：

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(); }
      };
    }
  };
}

有了这个，

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