Pattern match for (length)%code with before length - regex

Question

I have a pattern like x%c, where x is a single digit integer and c is an alphanumeric code of length x. % is just a token separator of length and code
For instance 2%74 is valid since 74 is of 2 digits. Similarly, 1%8 and 4%3232 are also valid.
I have tried regex of form ^([0-9])(%)([A-Z0-9]){\1}, where I am trying to put a limit on length by the value of group 1. It does not work apparently since the group is treated as a string, not a number.
If I change the above regex to ^([0-9])(%)([A-Z0-9]){2} it will work for 2%74 it is of no use since my length is to be limited controlled by the first group not a fixed digit.
I it is not possible by regex is there a better approach in java?

Answer 1

One way could be using 2 capture groups, and convert the first group to an int and count the characters for the second group.
\b(\d+)%(\d+)\b
\b Word boundary
(\d+) Capture group 1, match 1+ digits
% Match literally
(\d+) Capture group 2, match 1+ digits
\b Word boundary
Regex demo | Java demo
For example
String regex = "\\b(\\d+)%(\\d+)\\b";
String string = "2%74";
Pattern pattern = Pattern.compile(regex);
String strings[] = { "2%74", "1%8", "4%3232", "5%123456", "6%0" };
for (String s : strings) {
Matcher matcher = pattern.matcher(s);
if (matcher.find()) {
if (Integer.parseInt(matcher.group(1)) == matcher.group(2).length()) {
System.out.println("Match for " + s);
} else {
System.out.println("No match for " + s);
}
}
}
Output
Match for 2%74
Match for 1%8
Match for 4%3232
No match for 5%123456
No match for 6%0