java regex quantifiers

Question

I have a string like

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

I need a regex to give me the following output:

number0 foobar
number1 foofoo
number2 bar bar bar bar
number3 foobar

I have tried

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

but this gives

number0 foobar number1 foofoo number2 bar bar bar bar number3 foobar

Answer 1

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

Then you need to tell that to the regex engine:

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

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.

Explanation of my solution:

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.

Answer 2

Pattern pattern = Pattern.compile("\\w+\\d(\\s\\w+)\1*");
Matcher matcher = pattern.matcher(string);

while (matcher.find()) {
    System.out.println(matcher.group());
}

Answer 3

because .* is a greedy pattern. use .*? instead of .*

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

Answer 4

如果“foobar”只是一个例子而且你的意思是“任何单词”使用以下模式:( (number\\\\d+)\\s+(\\\\w+)

Answer 5

为什么不匹配number\\\\d+ ，查询匹配位置，自己进行字符串拆分？

Answer 6

(.*) part of your regex is greedy, therefore it eats everything from that point to the end of the string. Change to non-greedy variant: (.*)?

http://docs.oracle.com/javase/tutorial/essential/regex/quant.html