Java: Split last digit from the rest of the INT

16,625

Solution 1

int x = 1343248724;
int firstpart = x/10;
int secondpart = x%10;

System.out.println(firstpart);
System.out.println(secondpart);

Mod (%) gives you the remainder which will be the last digit if you mod 10 Divide will drop the last digit. That way you don't have to actually search the string.

Doing all the length stuff seems a little like overkill to me

oh and to get your param as a int I would use

Interger.decode(fiscalId);

Solution 2

This is how you can get the Integers:

Integer firstPart = Integer.valueOf(
                        fiscalId.substring(0,fiscalId.length()-1));
Integer secondPart = Integer.valueOf(
                        fiscalId.substring(fiscalId.length()-1));

but I'd suggest getting ints instead:

int firstPart = Integer.parseInt(fiscalId.substring(0,fiscalId.length()-1));
int secondPart = Integer.parseInt(fiscalId.substring(fiscalId.length()-1));
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16,625
BoDiE2003
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BoDiE2003

Updated on June 14, 2022

Comments

  • BoDiE2003
    BoDiE2003 almost 2 years

    I have the next code, which gets a number, I need to split that number in to parts, firstPart should be the whole number without the last digit, and the secondPart should be only the last digit of the original number.

    public Boolean verificationNumberFiscalId(String fiscalId) {
    
            // Trying to do a better wa yto do it
            Integer firstPart = fiscalId.length()-1;
            Integer secondPart = ??????????;
    
    
    
            // My old logic
            String NitValue = "";
            for (Integer i = 0; i < fiscalId.length()-1; i++) {
                NitValue = fiscalId.substring(i,1);
            }
    
            Integer actualValueLength = fiscalId.length()-1;        
            String digitoVerificador = fiscalId.substring(actualValueLength,1);
            return this.generateDigitoVerification(NitValue) == digitoVerificador;
        }
        /** 
         * @param nit
         * @return digitoVerificador
         * @comment: Does the math logic
         */
        public String generateDigitoVerification(String nit) {        
            Integer[] nums = { 3, 7, 13, 17, 19, 23, 29, 37, 41, 43, 47, 53, 59, 67, 71 };
    
            Integer sum = 0;
    
            String str = String.valueOf(nit);
            for (Integer i = str.length() - 1, j=0; i >= 0; i--, j++) {
                sum += Character.digit(str.charAt(i), 10) * nums[j];
            }
    
            Integer dv = (sum % 11) > 1 ? (11 - (sum % 11)) : (sum % 11);
            return dv.toString();
        }
    

    Could you suggest me a better way to do it please? Thank you!