Console input java. Parsing a string

Question

I have a console application where I need to input numbers until enter "x". Of course when I input "x" i will get a NumberFormatException.

How would I quit the program when entering "x" without getting an exception.

BufferedReader buf = new BufferedReader(new InputStreamReader(System.in));
    String s;
    int input;
    String name = args[0];
    SimpleDateFormat sdf = new SimpleDateFormat("HH:mm:ss");
    Date date = new Date();


    System.out.println("Good morning " + name + " Today's date is " + sdf.format(date));

    System.out.println("Please enter any number between 0 and 10");

    try
    {


        do
        {
        s = buf.readLine();

        input = Integer.parseInt(s);

        while(input <= 0 || input > 10)
        {
            System.out.println("Make sure about the correct input...between 0 and 10 please");
            s = buf.readLine();
            input = Integer.parseInt(s);
            System.out.println(input);
        }
        }while(s != "x");

Hayden · Accepted Answer · 2012-03-28 19:22:22Z

3

Just add a line

s = buf.readLine();
if ("x".equals(s)) break; // add this line

This does not guarantee that s is an integer, so you still need to catch an exception like andreas proposed

edited Mar 28, 2012 at 19:22

Hayden

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answered Mar 28, 2012 at 9:41

stefan bachert

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Comments

hmjd · Accepted Answer · 2012-03-28 09:51:43Z

3

Reorder the loop(s) to check if s is equal to x before Integer.parseInt(). Use String.equals() to compare strings, not == or !=.

As this is homework I will not post modified code.

EDIT:

Just to explain the reason for using String.equals():

From section 15.21.3 Reference Equality Operators == and != of the Java Language Specification 3.0:

While == may be used to compare references of type String, such an equality test determines whether or not the two operands refer to the same String object. The result is false if the operands are distinct String objects, even if they contain the same sequence of characters. The contents of two strings s and t can be tested for equality by the method invocation s.equals(t).

edited Mar 28, 2012 at 9:51

answered Mar 28, 2012 at 9:40

hmjd

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Comments

KV Prajapati · Accepted Answer · 2012-03-28 09:40:55Z

2

Wrap the code in try..catch block.

input=-1;
try{
  input = Integer.parseInt(s);
  while(input <= 0 || input > 10)
  {
   System.out.println("Make sure about the correct input...between 0 and 10 please");
   s = buf.readLine();
   input = Integer.parseInt(s);
   System.out.println(input);
   }
}catch(Exception ex) { }
....

answered Mar 28, 2012 at 9:40

KV Prajapati

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Comments

Andreas Dolk · Accepted Answer · 2012-03-28 09:41:20Z

2

Wrap the Integer.parseInt statement in a try/catch block:

try {
 input = Integer.parseInt(s);
} catch{NumberFormatException nfe) {
 System.out.println("Illegal input");
 // know you could do one of the following: (uncomment)
 // continue;        // would continue the while loop
 // break;           // would exit the while loop
 // System.exit(0);  // would exit the application
}

answered Mar 28, 2012 at 9:41

Andreas Dolk

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Comments

Darren · Accepted Answer · 2012-03-28 09:41:33Z

2

Change to a while loop and do

while (s != "x") {
  // your logic

}

or check if s != "x" before this line:

input = Integer.parseInt(s);

answered Mar 28, 2012 at 9:41

Darren

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Comments

Pau Kiat Wee · Accepted Answer · 2012-03-28 09:43:00Z

2

Catch the exception:

s = buf.readLine();
if("x".compareToIgnoreCase(s)) {
  // im quit
  return;
}

answered Mar 28, 2012 at 9:43

Pau Kiat Wee

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Comments

John Snow · Accepted Answer · 2012-03-28 09:40:18Z

1

Add a if statement before paring the string to a int:

    if(s.equals("x"))
       system.exit(1);

    else{
    input = Integer.parseInt(s);

            while(input <= 0 || input > 10)
            {
                System.out.println("Make sure about the correct input...between 0 and 10 please");
                s = buf.readLine();
                input = Integer.parseInt(s);
                System.out.println(input);
}

answered Mar 28, 2012 at 9:40

John Snow

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