How to get error to be displayed

Question

public void runMenu() { int x = 1; Scanner Option = new Scanner(System.in); int Choice = 0; do { try { System.out.println("Choose Option"); System.out.println(""); System.out.println("1: Create Account"); System.out.println("2: Check Account"); System.out.println("3: Take Action"); System.out.println("4: Exit"); System.out.println("Please choose"); Choice = Option.nextInt(); switch (Choice) //used switch statement instead of If else because more effective { case 1: CreateAccount(); break; //breaks iteration case 2: selectAccount(); break; case 3: break; } } catch (Exception e) { System.out.println("Cant do that"); } } while (Choice < 4); System.out.println("Bye"); }

The exception handling doesnt work e.g. if I enter say outide the range such as 5 it should say something like "Cant do that". Should I enter a if statement defining the condition if so how can I go about doing that? right now it just prints "Bye" if I enter wrong key.

Input outside of your range doesn't cause an exception. Exceptions are triggered by an input that causes a problem when processed.This might help you tutorialspoint.com/java/java_exceptions.htm — Ceelos
– Ceelos, Commented Feb 24, 2016 at 0:29
There is no part in your code that tells it to throw an exception if the input is an unexpected value. Maybe use the default: case of the switch. (Whether using exceptions here in the first place is correct is debatable) — Matti Virkkunen
– Matti Virkkunen, Commented Feb 24, 2016 at 0:31
What about a incorrect input e.g. if you set an if statement (Choice<4) can it be done then to display user an error and return back to menus? — Edafy
– Edafy, Commented Feb 24, 2016 at 0:33
"if I enter say outide the range such as 5 it should say something like "Cant do that"" <-- what exactly makes you think that the current code will do that? .nextInt() will try and grab any valid integer from the input. 5 is valid... — fge
– fge, Commented Feb 24, 2016 at 0:36

Moe · Accepted Answer · 2016-02-24 00:31:48Z

Your issue is due to the fact that your looping while Choice < 4.

You should change your loop to while(true); and then inside your switch case, create a case 4 which exits the program using System.exit(0)

Then include a default which will inform the user that they input an invalid value

public void runMenu() { int x = 1; Scanner Option = new Scanner (System.in); int Choice = 0; do{ try{ System.out.println("Choose Option"); System.out.println(""); System.out.println("1: Create Account"); System.out.println("2: Check Account"); System.out.println("3: Take Action"); System.out.println("4: Exit"); System.out.println("Please choose"); Choice= Option.nextInt(); switch (Choice) //used switch statement instead of If else because more effective { case 1: CreateAccount(); break; //breaks iteration case 2: selectAccount(); break; case 3: break; case 4: System.out.println("Bye"); System.exit(0); default: System.out.println("Invalid Entry"); break; } } catch (Exception e){ System.out.println("Cant do that"); } } while (true); }

sanky jain · Accepted Answer · 2016-02-24 00:40:51Z

Currently you are not throwing any exception.

If you want to throw an exception. Please create default case and throw exception from it

Made default case as below

default: System.out.println("Invalid Entry"); throw new Exception();

.nextInt() from Scanner can throw plenty of exceptions. And they are all unchecked to boot.

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