如何匹配这种模式？ （Java /正则表达式）

Question

I need to transform the string:

"%s blabla %s"

into:

"%1$s blabla %2$s"

My code is as follows:

Pattern pattern = Pattern.compile("%s");
Matcher tokenMatcher = pattern.matcher(value);
int index = 1;
while(tokenMatcher.find()){
    String replacement = "%"+String.valueOf(index++)+"\\$s";        
    value = tokenMatcher.replaceFirst(replacement);
    System.out.println(value);
}

The problem is that the program gets in an infinite loop and I don't understand why. Somehow %1$s is matched by %s

%1$s blabla %s
%2$s blabla %s
%3$s blabla %s
%4$s blabla %s
%5$s blabla %s
%6$s blabla %s
%7$s blabla %s
%8$s blabla %s
%9$s blabla %s
%10$s blabla %s
etc...

Any idea?

Answer 1

try resetting the tokenMatcher in the loop.

while(tokenMatcher.find()){
    String replacement = "%"+String.valueOf(index++)+"\\$s";        
    value = tokenMatcher.replaceFirst(replacement);
    tokenMatcher = pattern.matcher(value);
}

System.out.println(value);

Answer 2

You have to reset your matcher with the new value:

while (tokenMatcher.find()) {
  String replacement = "%" + String.valueOf(index++) + "\\$s";
  value = tokenMatcher.replaceFirst(replacement);
  tokenMatcher.reset(value); // reset
  System.out.println(value);
}

The reason is that replaceFirst() reset the matcher to the beginning but does not change the string it is currently matching, it's still contains the old string. You have to do that yourself to update the matcher.

Answer 3

while(matcher.find()){
    matcher.appendReplacement(stringBuffer, "%" + String.valueOf(index++) + "\\$s");
}
matcher.appendTail(stringBuffer);

http://tutorials.jenkov.com/java-regex/matcher.html#8

Answer 4

Try below code

 StringBuilder stringBuilder = new StringBuilder();
        int i = 1;

        for (String s : input.split("%s")) {
            stringBuilder.append(String.format("%s %d", s, "%"+i+++"$"));


       }

String newString = stringBuilder.toString();