How to get the last characters in a String in Java, regardless of String size.

Question

Hi,

I'm looking for a way to pull the last characters from a String, regardless of size. Lets take these strings into example:

"abcd: efg: 1006746"
"bhddy: nshhf36: 1006754"
"hfquv: nd: 5894254"

As you can see, completely random strings, but they have 7 numbers at the end. How would I be able to take those 7 numbers?

Edit:

I just realized that String[] string = s.split(": "); would work great here, as long as I call string[2] for the numbers and string[1] for anything in the middle.

Answer 1

How about:

String numbers = text.substring(text.length() - 7);

That assumes that there are 7 characters at the end, of course. It will throw an exception if you pass it "12345". You could address that this way:

String numbers = text.substring(Math.max(0, text.length() - 7));

or

String numbers = text.length() <= 7 ? text : text.substring(text.length() - 7);

Note that this still isn't doing any validation that the resulting string contains numbers - and it will still throw an exception if text is null.

Answer 2

This should work

 Integer i= Integer.parseInt(text.substring(text.length() - 7));

Answer 3

Lots of things you could do.

s.substring(s.lastIndexOf(':') + 1);

will get everything after the last colon.

s.substring(s.lastIndexOf(' ') + 1);

everything after the last space.

String numbers[] = s.split("[0-9]+");

splits off all sequences of digits; the last element of the numbers array is probably what you want.

Answer 4

I'd use either String.split or a regex:

---

Using String.split

String[] numberSplit = yourString.split(":") ; 
String numbers = numberSplit[ (numberSplit.length-1) ] ; //!! last array element

---

Using RegEx (requires import java.util.regex.*)

String numbers = "" ;
Matcher numberMatcher = Pattern.compile("[0-9]{7}").matcher(yourString) ;
    if( matcher.find() ) {
            numbers = matcher.group(0) ;
    }