如果用户输入String而不是Int，则可能出现异常

Question

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 1

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 2

将用于读取int的代码嵌入到try catch块中，每当输入错误的输入时，它将生成一个异常，然后在catch块中显示所需的任何消息

Answer 3

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 4

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 5

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 if a String is numeric in Java for more info

Answer 6

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!

Answer 7

You want an Exception to occur if the user enters a string such as 'abc' instead of an integer value then the InputMismatchException is right for you.

Let me give a basic example for you.

public static void main(String[] args)
    {
        Scanner ip = new Scanner(System.in);
        int a; 
        System.out.println("Enter Some Input");
        try{
            a = ip.nextInt();
        }
        catch(InputMismatchException msg){
            System.out.println("Input Mismatch Exception has occured " + msg.getMessage());
        }
    }