I was having some problem when trying to do a try catch for do while loop:
try{
do {
System.out.println("Enter your option: ");
choice = sc.nextInt();
switch (choice) {
case 1:
break;
case 2:
break;
case 3:
break;
case 4:
break;
case 5:
break;
}
} while (choice != 6);
}catch(InputMismatchException e){
System.out.println("Please enter option between 1-6.");
}
What I am trying to do for the do while loop is when user entered anything other than 6 which is terminate, it will keep prompting for user input. For each case, it will go to certain method.
Then, I tried to do a try catch for InputMismatchException because my Scanner is taking integer from user input. However, after I entered alphabet instead of integer, the program just terminated itself. I am trying to do like when user entered alphabet, it will keep on prompting user for correct input.
Any ideas? Thanks in advance.
I was thinking if I should make another do while to wrap the entire try catch?
Do like :
try {
choice = sc.nextInt();
} catch(InputMismatchException e){
System.out.println("Please enter option between 1-6.");
sc.next();
continue;
}
If user enters a invalid input it will go to the catch block and will continue the loop. Remove the outer try catch block. Its not required
To handle characters and and invalid numbers you could do something like this:
do {
System.out.println("Enter your option: ");
try{
choice = sc.nextInt();
catch(InputMismatchException e){
choice = 0;
sc.next();
}
switch (choice) {
case 1:
break;
case 2:
break;
case 3:
break;
case 4:
break;
case 5:
break;
default:
System.out.println("Please enter option between 1-6.");
break;
}
} while (choice != 6);
Related
I'm working on simple Calculator app on Java. When user enter 0(which returns info no math operator), i want to restart my function. But when I do it throws NoSuchElementException when debug pointer comes to int operationInput = sc.nextInt(); Here is the my whole code block. I tried try catch but it stucks. Maybe it cannot re-identify a variable because it doesn't quit of that code block.
static Object mathOperators() {
Scanner sc = new Scanner(System.in);
System.out.print("Please enter a number for choosing operation(if you don't know what operation equals to which number click 0): ");
int operationInput = sc.nextInt();
sc.close();
switch (operationInput) {
case 0:
System.out.println("1: Addition - 2: Subtraction - 3: Multiplication \n"
+ "4: Division - 5: Modulus");
return mathOperators();
case 1:
System.out.println(additionCalc());
break;
case 2:
System.out.println(substractionCalc());
break;
case 3:
System.out.println(multiplyCalc());
break;
case 4:
System.out.println(divisionCalc());
break;
case 5:
System.out.println(modulusCalc());
break;
default:
System.out.println("Please enter a valid number");
break;
}
return 0;
}
You can use only one Scanner(System.in) in your app. You must close the scanner before calling "break" in your switch cases.
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Hi guys so my program doesn't really catch any errors i.e. when I input a letter instead of a valid number it does catch the error but it doesn't return back to the menu , it just displays the statement. And when I use a number outside of the switch statement i.e. 5 it just loops back to the menu without displaying error. My code is below:
public void runMenu() {
Scanner Option = new Scanner (System.in);
int x = 1;
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:
Menu();
int choice = UserInput();
performAction(choice);
break;
case 4:
System.out.println("Thanks for using the application");
System.exit(0);
default:
throw new Exception();
// x=2; //if code doesn't run successfully then x !=2 leading to exception
}
}
The case 4 is not closed with a break therefore you never instantiate your exception !
You should have this at the end of your switch :
default:
throw new Exception();
break;
Also, you need to remove the return from catch section.
catch (Exception e){
System.err.println("Enter Correct Input");
return ;
}
for when you've entered a number other than 1-4 you should have a default option,
to re-run the questions after you've displayed an error message call runMenu()
for example,
case 4:
System.out.println("Thanks for using the application");
System.exit(0);
default:
println "Choose 1-4";
runMenu()
Hope this helps.
This really isn't a situation where an exception is normally thrown. If you want the program to just loop back to the menu if the user enters an invalid option, you should only have to add a default case to your while loop. You don't even need your x integer for that. You can try throwing a new Exception in the default case if you want.
import javax.swing.*;
import java.util.Arrays;
import java.util.Scanner;
public class runMenu {
public void runMenu() {
Scanner Option = new Scanner (System.in);
int Choice = 0;
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:
Menu();
int choice = UserInput();
performAction(choice);
break;
case 4:
System.out.println("Thanks for using the application");
System.exit(0);
// x=2; //if code doesn't run successfully then x !=2 leading to exception
throw new Exception();
break;
default:
System.out.println("Invalid option. Please try
again.");
throw new Exception();
runMenu();
}
}
}
}
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:
Menu();
int choice = UserInput();
performAction(choice);
break;
case 4:
System.out.println("Thanks for using the application");
System.exit(0);
default:
System.out.println("Invalid Entry");
throw new Exception();
}
} catch (Exception e) {
System.err.println("Enter Correct Input");
return;
}
} while (true);
}
I am trying to make it when users enter incorrect input type like a letter , the exception is caught and then returns back to the menus, right now it catches the exception but it doesnt stop running I have to force stop the program. So I added a return but that just displays the exception error and stops, how can I make it return back to the menus?
That is because you're returning from the method itself in the catch block.
And Do not throw exceptions like that. Just use some boolean to know if the choice is valid and loop until the choice is entered correctly.Prefer not to use while(true), instead rely on a boolean flag everytime like below,
public void runMenu() {
int x = 1;
Scanner Option = new Scanner (System.in);
int Choice = 0;
boolean isValidChoice = false;
do{
isValidChoice = false;
Choice = 0;
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");
if(Option.hasNextInt()){
Choice= Option.nextInt();
isValidChoice = true;
}
switch (Choice)
{
case 1:
CreateAccount();
break;
case 2:
selectAccount();
break;
case 3:
Menu();
int choice = UserInput();
performAction(choice);
break;
case 4:
System.out.println("Thanks for using the application");
System.exit(0);
default:
isValidChoice = false; //if invalid choice, then set flag to loop
System.out.println("Invalid Entry");
}
} while (!isValidChoice);
}
Move the "try {" after the "System.out.println("Please choose");" line.
you just need to remove the return in the catch. also just as a tip, you can get rid of the do while and just have a while loop, because the loop is never ending.
} catch (Exception e) {
System.err.println("Enter Correct Input");
}
Okay so I'm pretty sure this should work:
Create a boolean value outside of while loop that is holds if there was a valid input
boolean validInput = true;
In default set this value to false (meaning there is an invalid input)
default:
System.out.println("Invalid Entry");
validInput = false;
throw new Exception();
Make sure the catch statement is still in the do loop because the throw clause will halt normal execution and transition into exception execution. Next the while tester will test if there was a valid input
while(!validInput)
Lastly go up to the top of the do loop and set validInput to true. This will make it so that each time you clear the previous incorrect input.
This should work.
I'm basically trying to validate so that you can only enter an Integer. This is what I have at the moment, but if I type letters it goes through the switch and just leaves the result as blank.
I want it so that if anything other than an integer is entered it will go to default in the switch.
Any help would be great. Thanks!
while(loop && kb.hasNextInt())
{
choice = kb.nextInt();
switch(choice)
{
case 1 :
language = "FRENCH";
loop = false;
break;
case 2 :
language = "GERMAN";
loop = false;
break;
case 3 :
language = "SPANISH";
loop = false;
break;
default :
System.out.println("That is not a correct choice. Please try again!");
break;
}
}
If the next input is not an integer,
then .hasNextInt() will return false,
and therefore the loop will terminate early.
If you want to allow text input and respond to it,
then you need to read line by line, text instead of numbers,
and parse the line read with Integer.parseInt.
If the line cannot be parsed, you will get a NumberFormatException.
You can catch it, and handle appropriately.
while (loop && scanner.hasNextLine()) {
String line = scanner.nextLine();
try {
choice = Integer.parseInt(line);
} catch (NumberFormatException e) {
System.out.println("That is not an integer. Please try again!");
continue;
}
switch (choice) {
case 1:
language = "FRENCH";
loop = false;
break;
case 2:
language = "GERMAN";
loop = false;
break;
case 3:
language = "SPANISH";
loop = false;
break;
default:
System.out.println("That is not a correct choice. Please try again!");
break;
}
}
This is because a letter will cause your while(loop && kb.hasNextInt()) to be false. I suggest put an if statement with the hasNextInt() within the while loop.
Example (using a while loop instead of if statement to really try getting the number):
while(loop)
{
// validate int using while loop
while(!kb.hasNextInt())
{
System.out.println("you must enter a number! ");
kb.next();
}
choice = kb.nextInt();
switch(choice)
{
case 1 :
language = "FRENCH";
loop = false;
break;
case 2 :
language = "GERMAN";
loop = false;
break;
case 3 :
language = "SPANISH";
loop = false;
break;
}
}
System.out.println("Thank You " + studentID + " you have been registered for " + language);
This code will blow before it even begins if the user did not enter a number as the while required kb.hasNextInt() to be true (have a number) to even run.
What I do is that I usually put the validation around where I receive the input:
int choice;
Boolean retry = null;
while(retry == null) {
try{
String input = scanner.nextLine();
choice = Integer.parseInt(input);
retry = false;
}catch(NumberFormatException e){
System.out.println("Please enter a number from 1 to 4.");
}
}
switch(choice){
case 1:
// Do stuff
break;
case 2:
// Do stuff
break;
case 3:
// Do stuff
break;
case 4:
// Do stuff
break;
default:
System.out.println("Something went wrong!");
}
While writing code for my program I thought of testing the first part before moving on and writing the operations. Although I have the user input, but I want the options to be displayed after each operation (add, deltete..) is done untill the users presses exit. How do I modify my code to do it?
import java.io.*;
import java.util.*;
public class Records {
public static void main(String[] args) {
int choice;
do {
System.out.println("1.Add \n 2.Delete \n 3.Update \n 4.Show \n Exit");
//BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
//int choice;
System.out.println("Enter your Choice : ");
Scanner sc = new Scanner(System.in);
choice = sc.nextInt();
switch (choice) {
case 1:
System.out.println("Getting ready to Add a Record ");
//set();
break;
case 2:
System.out.println("Getting ready to Delete a Record ");
//delete();
break;
case 3:
System.out.println("Getting ready to Update a Record ");
//update();
break;
case 4:
System.out.println("Here is your record ");
//display();
break;
case 5:
System.out.println("Out we go.");
System.exit(0);
//exit();
break;
default:
System.out.println("Try again");
break;
}
} while ( choice > 5 || choice < 1 );
}
}
Simply change your while condition to:
} while ( choice > 0 && choice < 5 );
First make sure your scanner really has an int, use sc.hasNextInt() to validate the user entered a number. To end the do/while loop at "5.Exit", just have it like do{...}while(choice!=5). Code below is not tested.
import java.io.*;
import java.util.*;
public class Records {
public static void main(String[] args) {
int choice;
do {
System.out.println("1.Add \n 2.Delete \n 3.Update \n 4.Show \n 5.Exit");
System.out.println("Enter your Choice : ");
choice = -1;
Scanner sc = new Scanner(System.in);
// validate the next thing in your scanner is an int
// otherwise, sc.nextInt() might cause an exception
if (sc.hasNextInt()){
choice = sc.nextInt();
switch (choice) {
case 1: System.out.println("Getting ready to Add a Record ");
// ...
break;
case 2: System.out.println("Getting ready to Delete a Record ");
// ...
break;
case 3: System.out.println("Getting ready to Update a Record ");
// ...
break;
case 4: System.out.println("Here is your record ");
// ...
break;
case 5: System.out.println("Out we go.");
// skip System.exit(0), your main method ends
// automatically when you leave your do/while loop
break;
default: System.out.println("Try again");
break;
}
}
// if choice == 5 it ends, otherwise, starts over...
} while ( choice != 5 );
}
}
Although I have the user input, but I want the options to be displayed
after each operation (add, deltete..) is done untill the users presses
exit.
You can set int flag=0; and when user selects exit option set flag to 1 to tell loop to exit.As of now you are already breaking out for number > 5 or < 1 in default case so no need to put that condition in while.
int flag=0;//Declare outside the loop
do{
...
case 5: System.out.println("Out we go.");
flag=1;//Set flag to 1 if user enters 5
break;
...
} while ( flag!=1 );//Exit the loop when flag==1
//Or directly while ( choice!=5 );
Serious EDIT
As Java Programmer I should probably suggest you to use boolean primitive type for flagging.
boolean flag=true;//Declare outside the loop
do{
...
case 5: System.out.println("Out we go.");
flag=false;//Set flag to 1 if user enters 5
break;
...
} while (flag);//Exit the loop when flag==false
One More thing:
Surround code with try-catch to leave out invalid inputs and prompt again for input.
Most of the times it's not recommended to swallow the Exception.
do{
try{
....//Your Switch case
}catch(InputMismatchException e){}
} while (choice !=5);//But remove System.exti(0); from your switch statement
Simply make your while loop as:
} while ( choice!=0 );(The Wrong one)
Correction:
} while(choice!=0 && choice<=5)