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Home/ Questions/Q 7554969
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Editorial Team
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Editorial Team
Asked: May 30, 20262026-05-30T11:28:41+00:00 2026-05-30T11:28:41+00:00

I am working on this very simple method I know I am very close

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I am working on this very simple method I know I am very close to finish it, but I am missing a detail. I appreciate any help. Thank you.

/**
  Gets the first letter in this string.
  @return the FIRST LETTER, or "" if there are no letters.
  add1=AD3F add2=EF4G result=32SFB  (BUT THESE ARE RANDOM ONLY INTS AND CHARS)
*/
public String firstLetter()
{
    String line = add1+add2+result;

    for(int i=0; i<line.length(); i++){
    char ch=new Character(line.charAt(i));
    if(Character.isLetter(ch)){
        System.out.println("This is the first letter"+ch);
        return ch;
    }
    else
        System.out.println("No it is not a character: "+ch);
    return "";
    }
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1 Answer

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  1. Editorial Team
    Editorial Team
    2026-05-30T11:28:43+00:00Added an answer on May 30, 2026 at 11:28 am

    You need to move the else code outside the loop so that it is only executed when you have tested all of the characters:

    public class FirstLetter
    {
        public static void main(String[] args)
        {
            System.out.println(firstLetter());
        }
    
        public static String firstLetter()
        {
            String line = "AD3F" + "EF4G" + "32SFB";
    
            for (int i = 0; i < line.length(); i++)
            {
                char ch = line.charAt(i);
                if (Character.isLetter(ch))
                {
                    System.out.println("This is the first letter: " + ch);
                    return Character.toString(ch);
                }
    
            }
            System.out.println("No character found");
            return "";
        }
    }
    

    This kind of problem is more obvious when you format the code clearly.

    I have kept the return type as String as per your original, but see also Jon Skeet’s comments about changing that to char.

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