字符串中的Java缺点数不起作用

Question

I am trying to just find the number of consonants in a string. The following is the code I have for the method, but it keeps returning 0 when I run it. Am I using the ! wrong? Do I need to do it for each individual case? ie: !(ch == ('a')) || !(ch == ('o'))

public int numCons() {
    int i = 0;
    int length = quote.length();
    int con = 0;
    String string;

    for (i = 0; i < length; i++) {
        //string = quote.substring(i);
        char ch = quote.charAt(i);

        if (!(ch == ('a') || ch == ('e') || ch == ('i') 
                || ch == ('o') || ch == ('u') || ch == ('y') || ch == ('A') 
                || ch == ('E') || ch == ('I') || ch == ('O') || ch == ('U') || ch == ('Y'))) 
            if (Character.isLetter(i)) 
                con++;
        }
    return con;
}

Answer 1

i is the index you iterate over, not the character. You probably meant check

if (Character.isLetter(ch)) 

Answer 2

by the way if you're using Java 8+ then you can do this in one line with use of Stream API:

quote.chars().filter(c -> !"aeiou".contains(String.valueOf((char) c).toLowerCase())).count();

(there's probably an even more elegant way, this was just the first thing i wrote)