String.replaceAll变体

Question

Is there a quick way to replace all of some pattern occurrences with data derived from the matched pattern?

For example, if I wanted to replace all occurrences of a number within a string with the same number padded to fixed length with 0s.

In this case if the length is 4, then ab3cd5 would become ab0003cd0005 .

My idea was using a StringBuilder and 2 patterns: one would get all numbers and the other would get everything that is not a number, and appending the matches to the builder by the index the matches were found.

I think there might be something simpler.

Answer 1

You can probably achieve what you're after using appendReplacement and appendTail , something like this:

import java.util.regex.Pattern; import java.util.regex.Matcher;

String REGEX = "(\\d+)";
String INPUT = "abc3def45";
NumberFormat formatter = new DecimalFormat("0000");

Pattern p = Pattern.compile(REGEX);
Matcher m = p.matcher(INPUT); // get a matcher object
StringBuffer sb = new StringBuffer();
while(m.find()){
    m.appendReplacement(sb,formatter.format(Integer.parseInt(m.group(1))));
}

m.appendTail(sb);

String result = sb.toString();

Answer 2

If you know exactly how many zeros you want to pad before any single number, then something like this should work:

String text = "ab3cd5";
text = text.replaceAll("\\d","0000$0");
System.out.println(text);

Otherwise:

Pattern pattern = Pattern.compile("\\d+");
Matcher matcher = pattern.matcher(text);

StringBuffer result = new StringBuffer();
while(matcher.find()){
    matcher.appendReplacement(result, String.format("%04d", Integer.parseInt(matcher.group()))); 
}
matcher.appendTail(result);
System.out.println(result);

The format %04d means: an integer, padded by zero up to a length of 4.