java正则表达式量词 [英] java regex quantifiers

问题描述

我有一个字符串

String string = "number0 foobar number1 foofoo number2 bar bar bar bar number3 foobar";

我需要正则表达式给我以下输出：

I need a regex to give me the following output:

number0 foobar
number1 foofoo
number2 bar bar bar bar
number3 foobar

我试过

Pattern pattern = Pattern.compile("number\\d+(.*)(number\\d+)?");
Matcher matcher = pattern.matcher(string);
while (matcher.find()) {
    System.out.println(matcher.group());
}

但是这给出了

number0 foobar number1 foofoo number2 bar bar bar bar number3 foobar

推荐答案

所以你想要数字（+一个整数）后跟任何东西，直到下一个数字（或字符串结束），对吧？

So you want number (+ an integer) followed by anything until the next number (or end of string), right?

然后你需要告诉正则表达式引擎：

Then you need to tell that to the regex engine:

Pattern pattern = Pattern.compile("number\\d+(?:(?!number).)*");

在正则表达式中，。* 匹配尽可能多 - 直到字符串结尾的所有内容。另外，你创建了第二部分（number \\d +）？比赛本身的一部分。

In your regex, the .* matched as much as it could - everything until the end of the string. Also, you made the second part (number\\d+)? part of the match itself.

我的解决方案说明：

number    # Match "number"
\d+       # Match one of more digits
(?:       # Match...
 (?!      #  (as long as we're not right at the start of the text
  number  #   "number"
 )        #  )
 .        # any character
)*        # Repeat as needed.

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