How to get numeric position of alphabets in java?

String str = "abcdef";
char[] ch  = str.toCharArray();
for(char c : ch){
    int temp = (int)c;
    int temp_integer = 96; //for lower case
    if(temp<=122 & temp>=97)
        System.out.print(temp-temp_integer);
}

Output:

123456

@Shiki for Capital/UpperCase letters use the following code:

String str = "DEFGHI";
char[] ch  = str.toCharArray();
for(char c : ch){
    int temp = (int)c;
    int temp_integer = 64; //for upper case
    if(temp<=90 & temp>=65)
        System.out.print(temp-temp_integer);
}

Output:

456789

char letter;
for(int i=0; i<text.length(); i++)
{
    letter = text.charAt(i);
    if(letter>='A' && letter<='Z')
        System.out.println((int)letter - 'A'+1);
    if(letter>='a' && letter<= 'z')
        System.out.println((int)letter - 'a'+1);
}

Convert each character to its ASCII code, subtract the ASCII code for "a" and add 1. I'm deliberately leaving the code as an exercise.

This sounds like homework. If so, please tag it as such.

Also, this won't deal with upper case letters, since you didn't state any requirement to handle them, but if you need to then just lowercase the string before you start.

Oh, and this will only deal with the latin "a" through "z" characters without any accents, etc.

Another way to do this problem besides using ASCII conversions is the following:

String input = "abc".toLowerCase();
final static String alphabet = "abcdefghijklmnopqrstuvwxyz";
for(int i=0; i < input.length(); i++){
    System.out.print(alphabet.indexOf(input.charAt(i))+1);
}