Question

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!

Answer 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);