Java8-使用regex表达式从string中查找string列表[重复]

Question

private List<String> getDataOnRegx(String str) {
    Pattern filerRegx = Pattern.compile("\\(([a-zA-Z,0-9]*)\\s*\\),?");
    Matcher regexMatcher = filerRegx.matcher(str);
    return Stream.of(str).
            filter(s -> regexMatcher.find() && regexMatcher.group(1) != null).
            map(r -> new String(regexMatcher.group(1))).
            collect(Collectors.toList());

}

private List<String> getDataOnRegx(String str) {
Pattern regex = Pattern.compile("\\(([a-zA-Z,0-9]*)\\s*\\),?");
    Matcher regexMatcher = regex.matcher(str);
    List<String>dataList = new ArrayList<String>();
    while (regexMatcher.find()) {
          if (regexMatcher.group(1) != null) {
                dataList.add(regexMatcher.group(1));
          }
    }
    System.out.println(dataList);
    return dataList; 

}

谢谢，阿图尔

Answer 1

在java-9中可以通过扫描程序#findall:

    String test = "(A,1),(Y,4),(F,5)";
    Scanner sc = new Scanner(test);
    Pattern filerRegx = Pattern.compile("\\(([a-zA-Z,0-9]*)\\s*\\),?");

    List<String> results = sc.findAll(filerRegx)
            .map(mr -> mr.group(1))
            .filter(Objects::nonNull)
            .collect(Collectors.toList());

    System.out.println(results);

我不知道这个重复的答案怎么不能回答你的问题。你需要做一个非常简单的改变：

  static final class MatchItr extends Spliterators.AbstractSpliterator<String> {
    private final Matcher matcher;

    MatchItr(Matcher m) {
        super(m.regionEnd() - m.regionStart(), ORDERED | NONNULL);
        matcher = m;
    }

    @Override
    public boolean tryAdvance(Consumer<? super String> action) {
        if (!matcher.find()) {
            return false;
        }

        if (matcher.group(1) == null) {
            return false;
        }

        action.accept(matcher.group(1));
        return true;
    }
}

并使用它：

MatchItr mIter = new MatchItr(regexMatcher);
StreamSupport.stream(mIter, false)
            .forEach(System.out::println);