Search string for non-alphabetic characters

[turnergirl24]

Is there an easier way to search a string for non-alphabetic characters than this??

x = 0;
while (x != firstName.length()){
if ( firstName.charAt(x) == 'a' || firstName.charAt(x) == 'b' || firstName.charAt(x) == 'c' || firstName.charAt(x) == 'd' || firstName.charAt(x) == 'e' ||

firstName.charAt(x) == 'f' || firstName.charAt(x) == 'g' || firstName.charAt(x) == 'h' || firstName.charAt(x) == 'i' || firstName.charAt(x) == 'j' ||

firstName.charAt(x) == 'k' || firstName.charAt(x) == 'l' || firstName.charAt(x) == 'm' || firstName.charAt(x) == 'n' || firstName.charAt(x) == 'o' ||

firstName.charAt(x) == 'p' || firstName.charAt(x) == 'q' || firstName.charAt(x) == 'r' || firstName.charAt(x) == 's' || firstName.charAt(x) == 't' ||

firstName.charAt(x) == 'u' || firstName.charAt(x) == 'v' || firstName.charAt(x) == 'w' || firstName.charAt(x) == 'x' || firstName.charAt(x) == 'y' ||

firstName.charAt(x) == 'z' || firstName.charAt(x) == 'A' || firstName.charAt(x) == 'B' || firstName.charAt(x) == 'C' || firstName.charAt(x) == 'D' ||

firstName.charAt(x) == 'E' || firstName.charAt(x) == 'F' || firstName.charAt(x) == 'G' || firstName.charAt(x) == 'H' || firstName.charAt(x) == 'I' ||

firstName.charAt(x) == 'J' || firstName.charAt(x) == 'K' || firstName.charAt(x) == 'L' || firstName.charAt(x) == 'M' || firstName.charAt(x) == 'N' ||

firstName.charAt(x) == 'O' || firstName.charAt(x) == 'P' || firstName.charAt(x) == 'Q' || firstName.charAt(x) == 'R' || firstName.charAt(x) == 'S' ||

firstName.charAt(x) == 'T' || firstName.charAt(x) == 'U' || firstName.charAt(x) == 'V' || firstName.charAt(x) == 'W' || firstName.charAt(x) == 'X' ||

firstName.charAt(x) == 'Y' || firstName.charAt(x) == 'Z'){
}else { //
throw new nonAlphabeticException("Must only be alphabetic characters");

[Fubarable]

yikes!

Even that won't work. You could use regular expressions to solve this, but I have a feeling that this is for an intro course. Otherwise, you could simply have a testString
String testString = "abcdefghijklmnopqrstuvwxyz";

and test to see if each char in your string, changed to lower case, is contained in the testString. See the String API for more details on the String#contains(....) method.

[mrmatt1111]

String remainingJunk = firstName.replaceAll("[a-zA-Z]","");

[turnergirl24]

Thanks for your help!!

[rdtindsm]

While checking every character against the contents of the string will work, it's a rather brute force method and horribly inefficient. Consider in the worst case comparing "zzzzzz"
against "ABCD ..Zabcd...z"). You have to make 52 comparisons for every character.

An "is_alpha(char unknown_char) is easy to write. In english ascii, characters are coded in sequence [Cap letters], [Numbers and some others],[Low case letters]. IIRC correctly,
A is X40 and a is X60. You can check that yourself.

code
boolean IsAlpha(char a){
return ((a = 'A' && a <= 'Z') ||
(a >= 'a' && a = 'z')
}

Realizing that we're encouraged to not provide explicitly coded answers, it is so much easier to write the function in code than to explain in english. I think the pedegogy is the same if the OP reads the code. Besides, there are deliberate bugs that are easy to find if the code is understood (edit: and one that wasn't intentional). Can be extended to isAlphaNumeric(), isControlCharacter(), et al. You can revise the function to "IsNotAlpha", or simply use !IsAlpha(a)

This is also C code and I'm not being careful about using production code. In C, char would probably be an int in production code, but char is a java primitive, so this should be pretty close.

Anyway, there are only 4 comparisons instead of 52.

[Fubarable]

Anyway, there are only 4 comparisons instead of 52.

Point well taken.