Searching a String for an integer. Datatype mismatch?

Question

I'm trying to compare a 4 digit String using charAt(x) to an integer in order to find out how many of that integer there is in the string.

public static boolean tickets(int numbers) {

int a = (numbers/100000);
int b = (numbers/10000) - a*10;
int c = numbers - a*100000 - b*10000;
int x = 0;
int count = 0;
String s = String.valueOf(c);

if (numbers/100000 == 0 || numbers/100000 > 9 ) { 
  return false;                                  
}
if (numbers <= 0) {
  return false;
} 

while (x < 4) {
  if (s.charAt(x) == a ) {
    count = count + 1;
  }
  x = x + 1;
      }

if (count == b) {
  return true;
} else {
  return false;
}}

I'm not going to insult anyone's intelligence and say this isn't a homework problem: it is. The prompt is to input a number (if negative, or not 6 digits, return ' false '). To get a return of ' true ' the number has to have the first digit the second digit's number of times in the last for digits (what?!). For example, 121001 would return true since 1 occers 2 times in the last 4 digits.

Anyway, my problem is that even though s.charAt(x) will have the same value as a , at least when I print them out and check (ie they'll both print 9's or something), my count variable doesn't seem to budge. Do I have a datatype mismatch? What can I do to fix this? Or have I approached this problem the wrong way entirely?

Thanks in advance!

Answer 1

You have the data type mismatch. Character '0' does not equal to integer 0. One quick fix could be:

if (s.charAt(x) == a + '0' ) {
    count = count + 1;
}

Answer 2

Your if condition is always false, because the data types do not match...

Try and use this as your if statements condition:

 if(Integer.parseInt(s.charAt(x)+"") == a)

Answer 3

I would recommend reading through this: Convert from Integer to String .

Basically, you could create a new String variable called aString and set it to Integer.toString(a) .