How is Guava Splitter.onPattern(..).split（）不同于 String.split（..）？

regex split java guava

2022-09-01 10:01:46

我最近利用了一个前瞻性正则表达式的强大功能来拆分一个字符串：

"abc8".split("(?=\\d)|\\W")

如果打印到控制台，则返回：

[abc, 8]

对这个结果非常满意，我想把它转移到番石榴进行进一步的开发，看起来像这样：

Splitter.onPattern("(?=\\d)|\\W").split("abc8")

令我惊讶的是，输出更改为：

[abc]

为什么？

答案 1

你发现了一个错误！

System.out.println(s.split("abc82")); // [abc, 8]
System.out.println(s.split("abc8"));  // [abc]

这是用于实际拆分 s 的方法（Splitter.SplittingIterator：：computeNext）：SplitterString

@Override
protected String computeNext() {
  /*
   * The returned string will be from the end of the last match to the
   * beginning of the next one. nextStart is the start position of the
   * returned substring, while offset is the place to start looking for a
   * separator.
   */
  int nextStart = offset;
  while (offset != -1) {
    int start = nextStart;
    int end;

    int separatorPosition = separatorStart(offset);

    if (separatorPosition == -1) {
      end = toSplit.length();
      offset = -1;
    } else {
      end = separatorPosition;
      offset = separatorEnd(separatorPosition);
    }

    if (offset == nextStart) {
      /*
       * This occurs when some pattern has an empty match, even if it
       * doesn't match the empty string -- for example, if it requires
       * lookahead or the like. The offset must be increased to look for
       * separators beyond this point, without changing the start position
       * of the next returned substring -- so nextStart stays the same.
       */
      offset++;
      if (offset >= toSplit.length()) {
        offset = -1;
      }
      continue;
    }

    while (start < end && trimmer.matches(toSplit.charAt(start))) {
      start++;
    }
    while (end > start && trimmer.matches(toSplit.charAt(end - 1))) {
      end--;
    }

    if (omitEmptyStrings && start == end) {
      // Don't include the (unused) separator in next split string.
      nextStart = offset;
      continue;
    }

    if (limit == 1) {
      // The limit has been reached, return the rest of the string as the
      // final item.  This is tested after empty string removal so that
      // empty strings do not count towards the limit.
      end = toSplit.length();
      offset = -1;
      // Since we may have changed the end, we need to trim it again.
      while (end > start && trimmer.matches(toSplit.charAt(end - 1))) {
        end--;
      }
    } else {
      limit--;
    }

    return toSplit.subSequence(start, end).toString();
  }
  return endOfData();
}

感兴趣的领域是：

if (offset == nextStart) {
  /*
   * This occurs when some pattern has an empty match, even if it
   * doesn't match the empty string -- for example, if it requires
   * lookahead or the like. The offset must be increased to look for
   * separators beyond this point, without changing the start position
   * of the next returned substring -- so nextStart stays the same.
   */
  offset++;
  if (offset >= toSplit.length()) {
    offset = -1;
  }
  continue;
}

此逻辑非常有效，除非空匹配发生在 .如果空匹配确实发生在的末尾，它最终会跳过该字符。这部分应该是什么样子的（注意->）：StringString>=>

if (offset == nextStart) {
  /*
   * This occurs when some pattern has an empty match, even if it
   * doesn't match the empty string -- for example, if it requires
   * lookahead or the like. The offset must be increased to look for
   * separators beyond this point, without changing the start position
   * of the next returned substring -- so nextStart stays the same.
   */
  offset++;
  if (offset > toSplit.length()) {
    offset = -1;
  }
  continue;
}

答案 2

番石榴似乎有一个错误，当一个模式与一个空字符串匹配时。如果您尝试创建并打印出匹配的内容：SplitterMatcher

Pattern pattern = Pattern.compile("(?=\\d)|\\W");
Matcher matcher = pattern.matcher("abc8");
while (matcher.find()) {
    System.out.println(matcher.start() + "," + matcher.end());
}

您将获得输出，使其看起来与 .因此，它只是在那里分裂，只产生。3,38abc

您可以使用例如这似乎给出了正确的输出：Pattern#split(String)

Pattern.compile("(?=\\d)|\\W").split("abc8")

How is Guava Splitter.onPattern(..).split（） 不同于 String.split（..）？

How is Guava Splitter.onPattern(..).split（）不同于 String.split（..）？