java正则表达式量词

Question

我有一个字符串

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

我需要一个正则表达式给我以下输出:

number0 foobar
number1 foofoo
number2 bar bar bar bar
number3 foobar

我试过了

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

但这给了

number0 foobar number1 foofoo number2 bar bar bar bar number3 foobar

Answer 1

所以你想要number(+一个整数)后跟任何东西,直到下一个number(或字符串的结尾),对吗？

然后你需要告诉正则表达式引擎:

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

在你的正则表达式中,.*尽可能匹配 - 直到字符串结尾的所有内容.此外,你将第二部分(number\\d+)?作为比赛的一部分.

解释我的解决方案:

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.