Evaluate the Last 2 Digits of an Int?

Question

I need to make a program that reads hours in this format (934:9h34) or 1835 (18h35). How can I make my program print an error if somebody writes 966 (the 2 last digits over 59? (66>59)

Answer 1

Given a String str :

String str = getTheString();
String lastTwoDigits = str.length() > 2 ? str.substring(str.length() - 2) : str;
int result = 0;
try {
  result = Integer.parseInt(lastTwoDigits);
} catch (NumberFormatException e) {
  System.err.println("Cannot parse string!");
  System.exit(1);
}
if (result > 59) {
  System.err.println("Number was over 59!");
  System.exit(1);
}

By the way, System.err.println() just prints to standard error rather than standard output, and exit(1) exits the program with a failing error code.

Hope this helps!

Answer 2

Since the title asks "How to evaluate the 2 last digit of a Int?", we can assume that you already have the value in an int variable.

To examine the last 2 digits, calculate the remainder when dividing by 100, ie use the % remainder operator:

int value = /*assigned elsewhere*/;

int lastTwoDigits = value % 100;
if (lastTwoDigits > 59) {
    System.out.println("ERROR: Invalid value: " + value);
    // value is invalid
}

Of course, you should probably also validate that value is not negative.

If, however, a value of -934 is valid, and -966 is not, just eliminate the sign by calling Math.abs() :

int lastTwoDigits = Math.abs(value) % 100;
if (lastTwoDigits > 59) {
    System.out.println("ERROR: Invalid value: " + value);
    // value is invalid
}

Answer 3

This will work. Ensures that last 2 digits are <= 59.

      String[] test = { "934:9h34", "1835", "1994", "iwiwiwiw45", "18h45"
      };
      // match 2 digits at end of string
      Pattern p = Pattern.compile("(\\d\\d)$");
      for (String t : test) {
         Matcher m = p.matcher(t);
         if (m.find()) {
            // valid integer so convert and compare
            if (Integer.valueOf(m.group(1)) <= 59) {
               System.out.println("Passes test: " + t);
               continue;
            }
         }
         System.out.println("Fails test: " + t);
      }

Learn more about Java and regular expressions here .

Answer 4

This solution will parse the string first, then get the last two digits of the number through result % 100 .

private static void timeFormat(String text) {
    int result = 0;
    if (text.length() < 2) {
        System.err.println("String was too short");
        return;
    }
    try {
        result = Integer.parseInt(text);
    } catch (NumberFormatException e) {
        System.err.println("Failed to parse string");
    }
    if (result % 100 > 59) {
        System.err.println("Number was over 59");
    }
}

public static void main(String[] args) { 
    Scanner scan = new Scanner(System.in);
    timeFormat(scan.nextLine());
    scan.close();
}

Answer 5

First convert user given input into String

String hourInString = Integer.toString(userInput);

Then check if the input is valid or not. Minimum length of the input should be at least 3.

if (hourInString.length() < 3) {
    System.out.println("invalid input");
    System.exit(1);
}

Then retrieve the last two digit using substring

String lastTwoDigit = hourInString.substring(hourInString.length() - 2, 
                      hourInString.length());

Finally you can validate the number-

if (Integer.parseInt(lastTwoDigit) > 59) {
    System.out.println("Error");
}