Is there an opposite method for .charAt()?

Question

The remove.charAt() does not seem to work so I was wondering if there was something, where I input an int and it tells me what character is at that point? Again if anyone has any ideas it would be greatly appreciated. :-)

public static String removeYak(String remove)
{
    String fina1 = "";
    for(int y = 0; y < remove.length(); y++)
    {
        if(remove.toLowerCase().charAt(y) == "y")
        {
            if(remove.toLowerCase().charAt(y+1) == "a")
            {
                if(remove.toLowerCase().charAt(y+2) == "k")
                {
                    fina1 = str.substring(k+1);

                }
            }
        }
    }
    return fina1;
}

Your title and actual question don't line up. What are you trying to accomplish and what's your question? — Andrew Li, Oct 17 '16 at 21:29
Also - `charAt` returns a char. You have to compare with a *char* literal, not a String literal... — Andrew Li, Oct 17 '16 at 21:30
Wait never mind I just realized the answer is right under my nose, 'if(remove.toLowerCase().substring(y+1,y+2) == "a")'. Whoops, sorry guys... — Snare_Dragon, Oct 17 '16 at 21:37
And you call `toLowerCase()` potentially hundreds of times. Just remember `String lower = remove.toLowerCase();`. — chrylis -cautiouslyoptimistic-, Oct 17 '16 at 22:06

score 0 · Accepted Answer · answered Oct 17 '16 at 21:39

0

No, there isn't. However, you're using a correct method. The problem is in your if's. When you do something like:

String string = "abc";
char c = string.charAt(2);

You must compare character to a character, not to a string. In your if's, you write things like remove.toLowerCase().charAt(y) == "y", however, what you're looking for is remove.toLowerCase().charAt(y) == 'y'. Notice the apostrophe by the y you must use.

answered Oct 17 '16 at 21:39

leonz

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Thanks, that added a whole different perspective to what I was doing. Thanks for the help! – Snare_Dragon Oct 17 '16 at 21:44

Is there an opposite method for .charAt()?

1 Answers1