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.
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.
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.
lastIndexOf approach is that if there aren't any spaces, lastIndexOf will return -1, so you'll end up with the whole string (s.substring(0)).
This question is the top Google result for "Java String Right".
Surprisingly, no-one has yet mentioned Apache Commons StringUtils.right():
String numbers = org.apache.commons.lang.StringUtils.right( text, 7 );
This also handles the case where text is null, where many of the other answers would throw a NullPointerException.
n characters. For example, you could do StringUtils.right(Objects.toString(someObj), 7).This code works for me perfectly:
String numbers = text.substring(Math.max(0, text.length() - 7));
You can achieve it using this single line code :
String numbers = text.substring(text.length() - 7, text.length());
But be sure to catch Exception if the input string length is less than 7.
You can replace 7 with any number say N, if you want to get last 'N' characters.
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) ;
}
[0-9]{7}$ to make sure it matches the last 7 digits.
String inputstr = "abcd: efg: 1006746"
int startindex = inputstr.length() - 10;
String outputtendigitstr = inputstr.substring(startindex);
Make sure you check string length is more than 10.
org.apache.commons.lang3.StringUtils.substring(s, -7)
gives you the answer. It returns the input if it is shorter than 7, and null if s == null. It never throws an exception.
This should work
Integer i= Integer.parseInt(text.substring(text.length() - 7));
StringUtils.substringAfterLast("abcd: efg: 1006746", ": ") = "1006746";
As long as the format of the string is fixed you can use substringAfterLast.
E.g : "abcd: efg: 1006746" "bhddy: nshhf36: 1006754" "hfquv: nd: 5894254"
-Step 1: String firstString = "abcd: efg: 1006746";
-Step 2: String lastSevenNumber = firstString.substring(firstString.lastIndexOf(':'), firstString.length()).trim();
String.split(), but it is worth noting thats.split(": ")is going to compile a newjava.uitl.regex.Patternevery time you call it, then match your string with that pattern, creating a regexMatcherand anArrayListbefore theString[]that is returned. It will be relatively slow and will allocate far more than necessary to solve this problem. Whether this matters depends on the nature of your application. I generally avoidsplit()unless I really need it. (Note thatsplit()does not use regex if you split on a single character.)