How to extract last X digits from fixed Y digit number using RegEx?
input : 1234567890123456
((?=\d{16}$)\d{10}|) gives 1234567890But
((?=\d{16}$)\d{10}$|) does not give 7890123456basically, I want last 10 digits if it is compulsory 16 digit no else nothing.
How to extract last X digits from fixed Y digit number using RegEx?
.{X}$
where X is number of digits you want to extract
You can see from regex101 that the regex ^\d{6}(\d{10})$ will check if there is 16digit from start to end of the String (so check the length) and capture the last 10.
^ : start of the String\d{6} : first 6 digit(\d{10}) : capture 10 digit$ : end of the StringSo you can use it in Java with
String s = "1234567890123456";
Pattern p = Pattern.compile("^\\d{6}(\\d{10})$");
Matcher m = p.matcher(s);
if(m.matches()){
System.out.println(m.group(1));
}
This can be checked on this ideone
You just have to add an else clause to print that the String is not valid.
Looks like you're not using Java and cannot use capture groups.
For your tibco tool you can use this regex with validation of 16 digits to grab last 10 digits:
(?<=^\d{6})\d{10}$
RegEx Breakup:
(?<=^\d{6}) is a positive lookbehind to assert that we have 6 digits from start to previous position\d{10}$ matched 10 digits at the end of string