问题
The code below asks the user how many racers he/she would like.
while (true) { // loops forever until break
try { // checks code for exceptions
System.out.println("How many racers should" + " participate in the race?");
amountRacers = in.nextInt();
break; // if no exceptions breaks out of loop
}
catch (InputMismatchException e) { // if an exception appears prints message below
System.err.println("Please enter a number! " + e.getMessage());
continue; // continues to loop if exception is found
}
}
If a number is entered at amoutnRacers = in.nextInt();
the code breaks out of the loop and the rest of the program runs fine; however, when I enter something such as "awredsf" it should catch that exception, which it does. Instead of prompting the user again it loops continuously, which to me does not make sense.
The program prints like this when looping continuously:
How many racers should participate in the race? How many racers should participate in the race? How many racers should participate in the race? How many racers should participate in the race? How many racers should participate in the race? How many racers should participate in the race? How many racers should participate in the race?Please enter a number! null Please enter a number! null Please enter a number! null Please enter a number! null Please enter a number! null Please enter a number! null Please enter a number! null ...
I do not understand what is going on amountRacers = in.nextInt();
so why is the user not able to enter a number?
回答1:
Just add input.next()
once you catch InputMismatchException.
catch (InputMismatchException e) { //if an exception appears prints message below
System.err.println("Please enter a number! " + e.getMessage());
input.next(); // clear scanner wrong input
continue; // continues to loop if exception is found
}
You need to clear the wrong input, which scanner automatically does not.
回答2:
You may need to create a Scanner class for getting standard input streamed from the keyboard. You should have a statement somewhere in your code that creates an instance of a Scanner class like: Scanner in = new Scanner(System.in);
so the " in " variable in your statement: amountRacers = in.nextInt(); waits and scans for entered input from the keyboard and stores it.
回答3:
Today i solved this problem :-) This is my code. I think that i help
public int choice () throws Exception{
Scanner read = new Scanner(System.in));
System.out.println("Choose the option from the upper list");
int auxiliaryChoiceMenu = 5;
int auxiliaryVariable = -1;
boolean auxiliaryBoolean = false;
while (!auxiliaryBoolean) {
try {
auxiliaryVariable = read.nextInt();
read.nextLine();
} catch (Exception e) {
System.out.println("incorrect data, try again"+e);
read.nextLine();
continue;
}
if (auxiliaryVariable<0 || auxiliaryVariable>auxiliaryChoiceMenu){
System.out.println("incorrect data, try again");
} else {
auxiliaryBoolean = true;
}
choiceMenu = auxiliaryVariable;
}
return choiceMenu;
//choicemenu is a external variable
}
来源:https://stackoverflow.com/questions/24857070/try-catch-inside-while-loop