Count the occurrences of a letter in a string

Question

I am trying to write a for loop in Java that will count the occurrences of a letter in a string. The user will enter the letter to count and the string in which to search. This is a very basic code, and we have not gotten to arrays or much else yet. (I realize that I declared letter twice, but my brain is dead at this point) This is what I have tried so far and am having trouble with, any help is appreciated:

Ok I changed my code per suggestions, but now it is only reading the first word of my sentence?

import java.util.Scanner;

public class CountCharacters {
public static void main(String[] args) {
    Scanner in = new Scanner(System.in);

    char letter;
    String sentence = "";
    System.out.println("Enter a character for which to search");
    letter = in.next().charAt(0);
    System.out.println("Enter the string to search");
    sentence = in.next();

    int count = 0;
    for (int i = 0; i < sentence.length(); i++) {
        char ch = sentence.charAt(i);
        if (ch == letter) {
            count++;
        }
    }
    System.out.printf("There are %d occurrences of %s in %s", count,
            letter, sentence);

}
}

Answer 1

I see a couple of issues. First you have two variables with the same name.

Second your if condition check for the lenght of the sentence to be greater then 0 instead of checking for character equality.

Scanner in = new Scanner(System.in);

char inLetter = "";
String sentence = "";
System.out.println("Enter a character for which to search");
inLetter = in.next().charAt(0);
System.out.println("Enter the string to search");
sentence = in.next();

int letter = 0;
for (int i = 0; i < sentence.length(); i++) {
    char ch = sentence.charAt(i);
    if (inLetter == ch) {
        letter++;
    }
}

System.out.print(sentence.charAt(letter));

I would also strongly suggest to validate the input (which is not done in the example above) instead of just assuming you got 1 character from the first input and 1 sentence in the second.