I have some code in my java IDE and I believe that it has support for console. I have used java before and know the ins and outs. However, this problem has not occured with me before. My code looks like this:
Scanner scanner = new Scanner(System.in);
int choice = scanner.nextInt();
System.out.println("Choose a class.");
TimeUnit.SECONDS.sleep(10);
System.out.println("Press 1 for the a class.");
TimeUnit.SECONDS.sleep(10);
System.out.println("Press 2 for the b class.");
TimeUnit.SECONDS.sleep(10);
System.out.println("Press 3 for the c class.");
TimeUnit.SECONDS.sleep(10);
switch(choice) {
case 1:
playerClass.chosenClass = "a";
break;
case 2:
playerClass.chosenClass = "b";
break;
case 3:
playerClass.chosenClass = "c";
break;
default:
System.out.println("Null class. Please press 1-3 to choose a player class.");
}
}
I have all the imports, and the playerClass class does exist. The problem is, the "System.out.println(String)"'s do not work.
You're printing the prompt after accepting the input. And as you haven't seen the prompt yet, you probably aren't typing any input. So you never get past Scanner.nextInt().
Your code doesn't make sense.
What you might want to do is set your logic more like:
System.out.println("Press 1 for xx, 2 for xx, or 3 for xx");
if(scanner.hasNext()){
choice = scanner.Int();
}
if(choice == 1){
//do whatever;
}
else if(choice == 2){
//do whatever;
}
I might not be understanding your question right but I think you need to make sure you keep everything in order.
Related
Hey guys i have got a small weird problem here, i am asking the user to input their menu choice and depending on what they choose it calls a certain method.
I have used scanner.next() after some googling but for some reason only when i enter 1 or 2, i press enter and then press say 1 again and then it actually works. But what is weird that it calls options 3, 4, 5 and 6, immediately without me having to input the number twice.
I have tried with scanner.nextLine() after the scanner.nextInt() and that just leaves me having to put my option 1 or 2 in with no result.
while(exit == 0)
{
System.out.println("\n");
System.out.println("Menu 1: Display fullname of the user \n");
System.out.println("Menu 2: Display of user information \n");
System.out.println("Menu 3: Change password \n");
System.out.println("Menu 4: List all of users in the library full name\n");
System.out.println("Menu 5: Search for a book\n");
System.out.println("Press 6 to search for a books location in the library\n");
System.out.println("Press 0 to exit\n");
System.out.println("Enter choice: ");
int menuChoice = scanner.nextInt();
scanner.next();
if(menuChoice == 1)
{
displayUserFullName();
}
else if(menuChoice == 2)
{
displayUserInformation();
}
else if(menuChoice == 3)
{
menuForChangePassword();
}
else if(menuChoice == 4)
{
displayAllUserInSystem();
}
else if(menuChoice == 5)
{
searchBookByISBN();
}
else if(menuChoice == 6)
{
searchBookLocation();
}
else if(menuChoice == 0)
{
exit = 1;
}
}
Thank you in advance!
int menuChoice = scanner.nextInt();
scanner.next();
Read the javadoc for scanner. It waits for user input:
public String next(): [..] This method may block while waiting for input to scan
So in your program, you say: wait for user to type and int, then wait for user to type something.
Remove the scanner.next(); and it should work.
Scanner is a class parsing single tokens, like nextInt, nextDouble, nextToken (String). With corresponding testing methods: hasNextInt and so on.
All this parsing you do not need, so use nextLine for an entered line, or an other Reader class (InputStreamReader, BufferedReader).
Also you may utilize switch instead of if else if.
String menuChoice = scanner.nextLine();
switch (menuChoice) {
case "1":
displayUserFullName();
break;
case "2":
displayUserInformation();
break;
case "3":
menuForChangePassword();
break;
case "4":
displayAllUserInSystem();
break;
case "5":
searchBookByISBN();
break;
case "6":
searchBookLocation();
break;
case "0":
exit = 1;
break;
default:
System.out.printf("Unknown choice: '%s'%n", menuChoice);
}
menuChoice will contain the entire line, without line ending.
You might use an int with Integer.parseInt(menuChoice) but this would throw a NumberFormatException on wrong input, aborting your program. Scanner.nextInt would hang too, actually needing an hasNextInt().
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I have an exercise that using switch case. Assuming the code like
int choice;
do {
choice = //user input here;
switch (choice) {
case 1: //print something; break;
case 2: //print something; break;
case 3: //print something; break;
default: //print default; break;
}
} while(condition);
i want the user can only choose the case once. If they did choose case 1, they cannot choose that again.
Define a private bool variable for each one of your options and set it true if user called each one. Then check whether its true before running that option codes, again.
For example:
bool _case1=false;
int choice;
do{
choice = //user input here;
switch(choice){
case 1:
if(!_case1)
{
//print something;
_case1=true;
}
else
{
//Tell the user "You have selected this option once"
}
break;
//etc
}
}while(condition);
You could call methods for each case and have some bools that mark whether each case has already been chosen, use an if statement to check if the case has already been chosen and if so print a notice to pick another, otherwise perform the action.
You could also do this without separate methods but I am a fan of modular code. :)
I am assuming that a function that contains this switch case is called over and over again. If you want such a functionality, I would point to Set. Use the set to store the choices. If a set already contains a certain choice, take whatever action you deem fit.
Using a set is better than having a global array of booleans, since that could potentially be wasting a lot of memory, depending on the number of choices that you have.
Does that answer your question?
When user press the input
if the input is match case then
store this input in one List.
Next time you have to check this input.
If it belongs in List or not,
if belongs then case will not execute
if not belongs then it will execute the case.
Try in this way :
List<Integer> choiceList = new ArrayList<Integer>();
int choice;
do {
choice = //user input here;
if (choiceList.contains(choice)) {
continue;
}
choiceList.add(choice);
switch(choice) {
case 1: //print something; break;
case 2: //print something; break;
case 3: //print something; break;
default: //print default; break;
}
} while(condition);
As some of the other answers have said you can use a boolean to check if a case has been used or not. However, you can also use an int. The advantage of using an int is that you can specify how many times each case can be used. For example, the following code can only use each case once.
import java.util.Scanner;
public class Test {
public static void main(String[] args) {
Scanner user_input = new Scanner(System.in);
int choice;
int case1 = 0, case2 = 0, case3 = 0;
int case1lim = 1, case2lim = 1, case3lim = 1;
System.out.println("You may use may enter the number 1 " + case1lim + " times");
System.out.println("You may use may enter the number 2 " + case2lim + " times");
System.out.println("You may use may enter the number 3 " + case3lim + " times");
do {
System.out.println("Please enter a number between 1 and 3");
choice = user_input.nextInt();
if(choice == 1 && case1 < case1lim || choice == 2 && case2 < case2lim || choice == 3 && case3 < case3lim) {
switch(choice){
case 1: //print something;
case1++;
break;
case 2: //print something;
case2++;
break;
case 3: //print something;
case3++;
break;
default: //print default;
break;
}
} else {
System.out.println("Please pick another number since you have already used that case or you entered a bad value");
}
} while(true);
}
}
However if you change the values of the line
int case1lim = 1, case2lim = 1, case3lim = 1;
to
int case1lim = 2, case2lim = 1, case3lim = 1;
You can use the first case twice and all the other cases once.
My program contains a few options that the user can select via the input of a number which allows them to complete a specific task. Currently, my code is set up with if and else if loops to complete task if a certain number of input. However, at the minute the program terminates after one task. I want the user to be able to input another number to complete another task. I have tried surrounding the code with a while loop and an exit option to allow the user to escape the loop and end the program, but this is not working and results in a "java.util.NoSuchElementException". The program works fine without the while loop.
This is an example of the current code which hopefully conveys what I mean:
System.out.println("Enter one of the following commands:");
System.out.println("1 - something..");
System.out.println("2 - something else..");
System.out.println("3 - exit");
Scanner scanchoice = new Scanner(System.in);
System.out.println();
System.out.println("Enter \"1\", \"2\" or \"3\"");
int choiceentry = scanchoice.nextInt();
while (choiceentry != 3) {
if (choiceentry < 1 || choiceentry > 3) {
System.out.println("Enter \"1\", \"2\", \"3\" or \"4\"");
choiceentry = scanchoice.nextInt();
}
else if(choiceentry == 1) {
// ..do something
}
else if(choiceentry == 2) {
//..something else
}
else if(choiceentry == 3) {
//...exit program
}
}
So I want to get into this loop, and only exit to terminate the program. I'm hoping that the while loop would take the user back to a menu, allowing you to select another option, however this is not working. What is wrong with this code? And how can I implement this idea?
Thanks in advance!
Use Scanner#hasNextInt() before you call Scanner.nextInt() to get rid of the NoSuchElementException
if(scanchoice.hasNextInt())
choiceentry = scanchoice.nextInt();
hasNextInt() returns true only if the next token is a valid int
You can do like this
//set choiceentry to -1, this will make it to enter while loop
int choiceentry = -1
while(choiceentry < 1 || choiceentry > 3){
System.out.println("Enter \"1\", \"2\", \"3\" or \"4\"");
if(scanchoice.hasNextInt())
choiceentry = scanchoice.nextInt();
}
switch(choiceentry){
case 1:
//do logic
break;
case 2:
//do logic
break;
case 3:
//do logic
break;
}
I have changed it to use switch statements, since they come handy in getting input data
You are only asking the user to pick another menu item if choice is < 1 or > 3
you have to set this code in an else statement`:
while (choiceentry != 3) {
else if(choiceentry == 1) {
// ..do something
}
else if(choiceentry == 2) {
//..something else
}
else if(choiceentry == 3) {
//...exit program
}
else{
System.out.println("Enter \"1\", \"2\", \"3\" or \"4\"");
choiceentry = scanchoice.nextInt();
}
}
If you want your program to continue prompting the user to select a task you'll need to move that prompt as well as your nextInt() call to somewhere inside your loop yet outside of an if statement so that it will always be invoked on each iteration.
As Mr Phi suggested in the comments, a switch statement would be a better alternative to your current if-else structure. It'll make your code cleaner to read and a default case is pretty nice for catching unexpected values.
I'd also add that a do-while might be more suitable for this task. This way you won't need to code your prompt for a choice twice.
int choiceentry;
do {
System.out.println("Enter \"1\", \"2\" or \"3\"");
choiceentry = scanchoice.nextInt();
switch (choiceentry)
{
case 1:
// do something
break;
case 2:
// ..something else
break;
case 3:
// .. exit program
break;
default:
System.out.println("Choice must be a value between 1 and 3.");
}
} while (choiceentry != 3);
I'm having a slight problem.
I have a menu asking to:
reroll
get val
show max
show min
when the user chooses an option I want it to do one of them THEN re ask the menu in a sort of inifinite loop:
code:
import java.io.InputStream;
import java.util.Scanner;
class RecordDice {
public static void main(String[] args){
int dSides, Sides, Choice;
int max, min;
Scanner s = new Scanner(System.in);
Scanner c = new Scanner(System.in);
System.out.println("How many sides should the dice have?");
Sides = s.nextInt();
if(Sides == 4 || Sides == 6 || Sides == 12 || Sides == 20 || Sides == 100){
System.out.println("Please make a choice:\n" +
"1 - reroll the dice\n" +
"2 - get the value\n" +
"3 - show the maximum\n" +
"4 - show the minimum");
} else {
System.exit(-1);
}
Dice2 d = new Dice2(Sides);
int Choice = c.nextInt();
int Value = d.getValue();
switch(Choice){
case 1:
System.out.println();
d.reroll();
break;
case 2:
System.out.println("The current value is " + Value);
break;
case 3:
System.out.println("The maximum is " );
break;
case 4:
System.out.println("The minimun is ");
break;
}
}
}
Would putting the menu in a method and just calling the method every time a option is picked?
You can use a while loop to keep displaying it.
boolean keepGoing = true;
While(keepGoing)
{
//your code
}
Then to end it ask the user if they want to end it an set the boolean to false.
Add "5 - quit" to your menu.
Create a boolean, something like exit, initialized to false.
Add case 5: exit = true; break;
Then wrap the whole thing in while(!exit)
boolean exit = false;
while(!exit) {
//all the code you already have, starting with:
System.out.println("How many sides should the dice have?");
//and ending with the switch statement
//Plus the addition to the menu and addition to the switch statement
}
Ordinarily, I would do something like:
while(true) {
//do stuff
if(someExitCondition) {
break;
}
}
But seeing how as you're handling your user input with a switch statement, my above suggested method seems to be the cleanest way of handling it in this scenario.
Wrap it all in a do-while loop.
boolean userWantsToQuit = false;
do {
// code
// evaluate userWantsToQuit…
} while (!userWantsToQuit);
boolean keepGoing=true;
while(keepGoing)
{
//user input
if(user input to exit)
{
keepGoing=false;
}
}
or
while(true)
{
//user input
if(user input to exit)
{
break;
}
}
Assuming selection of dice sides you will allow only once, put code below that in do while loop.
You may prompt user "Do you wish to continue" after your switch block.
Get that value scanned
Condition in while loop will be something list while("YES".equals(userInput)).. assuming user will input YES or NO strings.
This is a small part of my program that I am working on. I'm trying to check if the user enters the correct number.
They have five choices to choose from so they can either hit 1, 2, 3, 4, or 5. Then press enter.
So I want to check to make sure the user doesn't type anything in < 1 or > 5. I got that part to work... But I just want to know if there is a easier way to do it then from what I did in code below.
The next part is that I also want to make sure the user doesn't type in letters. like "gfgfadggdagdsg" for a choice.
Here is my code of the part I am working on....
public void businessAccount()
{
int selection;
System.out.println("\nATM main menu:");
System.out.println("1 - View account balance");
System.out.println("2 - Withdraw funds");
System.out.println("3 - Add funds");
System.out.println("4 - Back to Account Menu");
System.out.println("5 - Terminate transaction");
System.out.print("Choice: ");
selection = input.nextInt();
if (selection > 5){
System.out.println("Invalid choice.");
businessAccount();
}
else if (selection < 1){
System.out.println("Invalid choice.");
businessAccount();
}
else {
switch(selection)
{
case 1:
viewAccountInfo3();
break;
case 2:
withdraw3();
break;
case 3:
addFunds3();
break;
case 4:
AccountMain.selectAccount();
break;
case 5:
System.out.println("Thank you for using this ATM!!! goodbye");
}
}
}
You may get rid of checking < 1 and > 5 by adding a default case.
try{
selection = input.nextInt();
switch(selection){
case 1:
viewAccountInfo3();
break;
case 2:
withdraw3();
break;
case 3:
addFunds3();
break;
case 4:
AccountMain.selectAccount();
break;
case 5:
System.out.println("Thank you for using this ATM!!! goodbye");
break;
default:
System.out.println("Invalid choice.");
businessAccount();
}
}catch(InputMismatchException e){
//do whatever you wanted to do in case input is not an int
}
Using BufferedReader you can do something like this:
InputStreamReader isr = new InputStreamReader(System.in);
BufferedReader br = new BufferedReader(isr);
String s = br.readLine();
int selection = 0;
try{
selection = Integer.parseInt(s);
if(selection > 5 || selection < 1){
System.out.println("Invalid choice.");
businessAccount();
}else{
// your switch code here
}
// you can use #Nishant's switch code here. it is obviously better: using switch's default case.
}catch(NumberFormatException ex){
// throw new Exception("This is invalid input"); // or something like that..
System.out.println("Invalid choice.");
businessAccount();
}
Hope that helps.
Note: you must import java.lang.NumberFormatException import java.io.InputStreamReader and import java.io.BufferedReader
Use the switch case it's better and more speed the if statement when you check selection from a Specific.
an alternative would to use regular expressions to get it work.
Say you have a string x then
String x = "something";
if(x.matches("regex")){
}
Another way to do this is surround with try catch.