substring to just get the last character

I have a variable called "last" that only contains numbers. I want to retrieve the last number in the series and that changes so I can't use a set number. I have tried using charAt(n-1) but that gives me a funky number that does do what I want. What I would like to see is if the last digit is 9, I want to see 9. How would be the best way to go about this?

If you first initialize n with:
n = last.length()
then your proposed solution should work.

That's what I thought but it doesn't seem to be working that way. Here is my code:

What I get for number is something like "52".

Try:

Yeah this works:

Originally posted by Candy Bortniker:
That's what I thought but it doesn't seem to be working that way. Here is my code:
int n = ltc.getText().length(); char last = ltc.getText().charAt(n-1); number = (long)last;
What I get for number is something like "52".

This is because last is the char representation of the digit. When you cast it to a long, you get the Unicode value of the character, which is obviously not what you expected.
To get the result you want, you can use the above suggestion:
number = (long)(last = '0');
HTH
Layne

I should note that this is only useful if you really want the int (or long) representation of the digit. If you are simply printing it out, you shouldn't waste your time and use the char representation instead.

I had to do a similar thing - check to see if the last character in a string was "s", if so, take it off:
//
// if myTimes is 1, and myTimesUnitsDesc ends in "s", strip it off
//
if (myTimes.intValue() == 1) {
int myLength = myTimesUnitsDesc.length();
if (myTimesUnitsDesc.charAt(myLength-1) == 's') {
return myTimesUnitsDesc.substring(0,myLength-1);
}
hope that helps

There's more than one way to skin a cat

:

Another approach that will have the bit-twiddlers* climbing the walls:
*bit-twiddler--someone willing to endure exercises in mental torture to shave a few microseconds from the execution time of an algorithm

Or if you require there be at least one character in the string besides the s (otherwise you want to leave that s alone), just change the * in the pattern string above to a +.
sev