Possible exception if User enters String instead of Int

Possible exception if User enters String instead of Int

I am just playing with Java.I'm trying to force my program to only accept 3 digit numbers. I believe I have successfully done this using a while loop (please correct me if I'm wrong). But how do I go about printing an error statement if the user enters a string. eg: "abc".

My code:

    import java.util.Scanner;      public class DigitSum {        public static void main(String[] args) {        Scanner newScan = new Scanner(System.in);            System.out.println("Enter a 3 digit number: ");          int digit = newScan.nextInt();            while(digit > 1000 || digit < 100)              {                          System.out.println("Error! Please enter a 3 digit number: ");               digit = newScan.nextInt();              }            System.out.println(digit);         }      }  

Answer by JonH for Possible exception if User enters String instead of Int

When you grab the input or pull the input string run through parseInt. This will in fact throw an exception if yourString is not an Integer:

Integer.parseInt(yourString)

And if it throws an exception you know its not a valid input so at this point you can display an error message. Here are the docs on parseInt:

http://docs.oracle.com/javase/1.4.2/docs/api/java/lang/Integer.html#parseInt(java.lang.String)

Answer by Software Sainath for Possible exception if User enters String instead of Int

Embed the code for Reading the int in try catch block it will generate an exception whenever wrong input is entered then display whatever message you want in catch block

Answer by MaVRoSCy for Possible exception if User enters String instead of Int

You can check if a String is of numeric value in the following ways :

1) Using a try/Catch Block

try    {      double d = Double.parseDouble(str);    }catch(NumberFormatException nfe)  {    System.out.println("error");  }    

2) Using regex

if (!str.matches("-?\\d+(\\.\\d+)?")){    System.out.println("error");  }  

3) Using NumberFormat Class

NumberFormat formatter = NumberFormat.getInstance();  ParsePosition pos = new ParsePosition(0);  formatter.parse(str, pos);  if(str.length() != pos.getIndex()){    System.out.println("error");  }  

4) Using the Char.isDigit()

for (char c : str.toCharArray())  {      if (!Character.isDigit(c)){        System.out.println("error");      }  }  

You can see How to check a String is a numeric type in java for more info

Answer by sakthisundar for Possible exception if User enters String instead of Int

Here nextInt method itself throws an InputMismatchException if the input is wrong.

try {    digit = newScan.nextInt()   } catch (InputMismatchException e) {    e.printStackTrace();    System.err.println("Entered value is not an integer");  }  

This should do.

Answer by Ruzia for Possible exception if User enters String instead of Int

How about this?

public class Sample {      public static void main (String[] args) {          Scanner newScan = new Scanner (System.in);            System.out.println ("Enter a 3 digit number: ");          String line = newScan.nextLine ();          int digit;          while (true) {              if (line.length () == 3) {                  try {                      digit = Integer.parseInt (line);                      break;                  }                  catch (NumberFormatException e) {                      // do nothing.                  }              }                System.out.println ("Error!(" + line + ") Please enter a 3 digit number: ");              line = newScan.nextLine ();          }            System.out.println (digit);      }  }  

regexp version:

public class Sample {      public static void main (String[] args) {          Scanner newScan = new Scanner (System.in);            System.out.println ("Enter a 3 digit number: ");          String line = newScan.nextLine ();          int digit;            while (true) {              if (Pattern.matches ("\\d{3}+", line)) {                  digit = Integer.parseInt (line);                  break;              }                System.out.println ("Error!(" + line + ") Please enter a 3 digit number: ");              line = newScan.nextLine ();          }            System.out.println (digit);      }  }  

Answer by anonymous for Possible exception if User enters String instead of Int

How I would do it is using an if statement. The if statement should be like this:

if(input.hasNextInt()){      // code that you want executed if the input is an integer goes in here      } else {     System.out.println ("Error message goes here. Here you can tell them that you want them to enter an integer and not a string.");  }  

Note: If you want them to enter a string rather than an integer, change the condition for the if statement to input.hasNextLine() rather than input.hasNextInt() .

Second Note: input is what I named my Scanner. If you name yours pancakes then you should type pancakes.hasNextInt() or pancakes.hasNextLine().

Hope I helped and good luck!

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