Make the scanner.nextInt() read a "q" as quit instead of reading it as a wrong input

Question

 int rollnumber;
 do {
     try {
         System.out.println("how many times");
         rollnumber = scanner.nextInt();

         nigh=2;

        if (Integer.toString(rollnumber).equals("q")  )   {
               System.exit(0);
         }   
     } catch (Exception e){
        System.out.println("invalid. re-enter");
        scanner.nextLine()';    
    }
}while (nigh==1);

Basically, I'm trying to make a "q" (quit) option but it just keeps jumping to the catch statement when I press q. Help me make it so that the "scanner.nextInt()" reads the q as the actual quit command.

You need to read from scanner.nextChar() check whether it's quit command and then if not convert it to int. — topr, Aug 15 '14 at 12:15
Nesting()accepts int so when you pass char then it will go to catch and its obvious — SpringLearner, Aug 15 '14 at 12:15
[this](http://stackoverflow.com/questions/13942701/take-a-char-input-from-the-scanner/13942707#13942707) may help :) — Reimeus, Aug 15 '14 at 12:45

Gabriel Negut · Answer 1 · 2014-08-15T12:22:48.117

3

You are trying to read an int from the console. If you enter q, nextInt attempts to convert q to an integer, fails with a NumberFormatException and the code jumps to the catch block.

You should read with next or nextLine and convert the input to an int.

edited Aug 15 '14 at 12:22

answered Aug 15 '14 at 12:17

Gabriel Negut

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There is no such method `nextChar` in `Scanner` – Ruchira Gayan Ranaweera Aug 15 '14 at 12:21
Whoops, my bad. Corrected. – Gabriel Negut Aug 15 '14 at 12:22
I made a: String stringy=scanner.next(); rollnumber=Integer.parseInt(stringy); and then tried to: if (rollnumber.equals("q") ) {System.exit(0);} however, the error is int cannot be deferenced. Im not sure what that means and what im doing wrong still – Boy Boy Aug 15 '14 at 13:00
You should check for 'q' BEFORE casting to int. – Gabriel Negut Aug 16 '14 at 07:14

Ruchira Gayan Ranaweera · Answer 2 · 2014-08-15T12:24:06.743

0

You can use this way. Read String not int.You can use nextLine()

Then

rollnumber = scanner.nextLine(); // read next line
if("q".equalsIgnoreCase(rollnumber)) // if it is q then if condition matched

edited Aug 15 '14 at 12:24

answered Aug 15 '14 at 12:17

Ruchira Gayan Ranaweera

32,406
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score 0 · Answer 3 · answered Aug 15 '14 at 12:21

The problem is that the .nextInt will only take in an actual int. You could use the .next() method instead of the .nextInt() method and then check to see if it is a "q". If it's not, you could then check if it parses to an Int and if not, use the code in the catch block.

String rollnumber;
try {
    System.out.println("how many times");
    rollnumber = scanner.next();

    nigh=2;

    if (rollnumber.equals("q")  )   {
        System.exit(0);
    } else {
        int num = Integer.parseInt(rollnumber);
    }
} catch (Exception e){
    System.out.println("invalid. re-enter");
    scanner.nextLine()';    
}

score -2 · Answer 4 · answered Aug 15 '14 at 12:16

Change your line to

int rollnumber;
nigh = 1; // added
do {  
 try {       
     System.out.println("how many times");
     String response = scanner.nextLine(); // changed

     rollnumber = scanner.nextInt();     
     nigh=2;
    if(if("q".equalsIgnoreCase(response)){){  // changed
       System.exit(0);
     } else {
             rollnumber = scanner.nextInt();     
             nigh=2;
     }
  }
 catch (Exception e){
    System.out.println("invalid. re-enter");
    scanner.nextLine();    
   }     
 }while (nigh==1);

Make the scanner.nextInt() read a "q" as quit instead of reading it as a wrong input

4 Answers4