创建正则表达式匹配数组 [英] Create array of regex matches

问题描述

在Java中我试图将所有正则表达式匹配返回到一个数组，但似乎你只能检查模式是否匹配（boolean）。有人可以帮我使用正则表达式匹配来形成一个匹配给定字符串中的正则表达式的所有字符串的数组吗？
谢谢！

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). Can someone help me use a regex match to form an array of all string matching a regex expression in a given string? Thanks!

推荐答案

（ 4castle的答案比下面的更好，如果你可以假设Java> = 9）

(4castle's answer is better than the below if you can assume Java >= 9)

你需要创建一个匹配器并使用迭代地找到匹配。

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());
 }

此后， allMatches 包含匹配项，如果您确实需要，可以使用 allMatches.toArray（new String [0]）来获取数组。

After this, allMatches contains the matches, and you can use allMatches.toArray(new String[0]) to get an array if you really need one.

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

You can also use MatchResult to write helper functions to loop over matches since Matcher.toMatchResult() returns a snapshot of the current group state.

例如，您可以编写一个惰性迭代器让你做

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.
}

做类似的事情：

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;
        }

        /** 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());
}

收益率

a at 0
b at 1
a at 3
c at 4
a at 5
a at 7
b at 8
a at 10

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