java外接程序只需编号

Question

我有一个样本字符串：

Today 2014 g. and i need buy 4135 g. coca-cola and 632.35 g bread and 7,3 g. salt. Notes by 1996

备注：

1. g可以是两端有点的，也可以是没有点的。
2. 2014 g.这是一年，可能是格式为20 or g.或19 or g。仅限

需要：

但我需要提取4135克和632.35克和7,3克只！

如果我们找到20 at (结尾有或没有g. )或19 at (结尾有或没有g. )--这就成了冰屋！(不要抽离！)

请帮助我，请为regex字符串(对于java)

Answer 1

这个regex可以帮你做到这一点：

\b(?!19|20)(\d+(?:\.\d+)?)\s+g\.?\b

Ie，在java正则表达式中：

private static final Pattern PATTERN
    = Pattern.compile("\\b(?!19|20)(\\d+(?:\\.\\d+)?)\\s+g\\.?\\b");

从输入中创建一个Matcher，使用.find()循环并为每个匹配提取.group(1)：

final Matcher m = PATTERN.matcher(input);

while (m.find())
    System.out.println(m.group(1));

正则表达式的分解：

\b            # Find a position where we have a word limit, then
(?!19|20)     # find a position where we don't have "19" or "20" following, then
(             # begin capturing group
  \d+         # one or more digits, followed by
  (?:         # begin non capturing group
    \.\d+     # one dot, followed by one or more digits
  )?          # end none capturing group, zero or one time,
)             # end capturing group, followed by
\s+           # one or more spaces, followed by
g\.?          # "g", then a literal dot, zero or one time, followed by
\b            # a word anchor again