We Keep Coding

Java, How do I validate input when using scanner?

I am currently working on Java code. Basically, the int input works. However, if I type in a character, the whole system crashes. My question is as to what needs to be changed in the below code in order for the user to receive a message stating that only an int is the valid input, and to try again if they input a character.
do {
System.out.println("How many players would like to participate in this game?\t(2-4 players)");
numberOfPlayers = in.nextInt();
} while(in.hasNextInt());
numberOfPlayers = in.nextInt();

I personally prefer to use a while loop for this sort of thing rather than the do/while. Not that there is anything wrong with the do/while, I just feel it's more readable to use the while loop.
I agree with others here, accept String digits from the User instead of Integer. In my opinion it saves you other possible problems down the road and you have no need to purposely apply a try/catch mechanism should the User supply an invalid entry. It also allows you to easily apply a mechanism to quit the application which, again IMHO, should be made available to all Console app's.
You've got your answer for carrying out the task using a do/while loop but I would like to show you another way to do this sort of thing:
Scanner in = new Scanner(System.in);
String ls = System.lineSeparator();
int numberOfPlayers = 0;
String userInput = "";
while (userInput.equals("")) {
// The Prompt to User...
System.out.print("How many players would like to participate in this game?" + ls
+ "2 to 4 players only (q to quit): --> ");
userInput = in.nextLine();
// Did the User enter: q, quit (regardless of letter case)
if (userInput.toLowerCase().charAt(0) == 'q') {
// No, the User didn't...
System.out.println(ls + "Quiting Game - Bye Bye.");
System.exit(0); // Close (exit) the application.
}
/* Did the User supply a string representation of a numerical
digit consiting of either 2, 3, or 4. */
if (!userInput.matches("[234]")) {
// No, the User didn't...
System.out.println("Invalid input! You must supply a number from 2 to 4 "
+ "(inclusive)." + ls + "Try again..." + ls);
userInput = "";
continue; // Loop again.
}
// Convert numerical string digit to an Ingeger value.
numberOfPlayers = Integer.parseInt(userInput);
}
System.out.println(ls + "The Number of players you provided is: --> "
+ numberOfPlayers);
You will notice that the Scanner#nextLine() method is used to accept User input as a String. This now means that we need to validate the fact that a string representation of a Integer numerical digit (2 to 4 inclusive) was supplied by that User. To do this you will notice that I used the String#matches() method along with a small Regular Expression (RegEx) which consists of the following string: "[234]". What this does in conjunction with the String#matches() method is it checks to see if the string value in the userInput variable contains either a single "2", a single "3", or a single "4". Anything else other than any one of those three digits will display this message:
Invalid input! You must supply a number from 2 to 4 (inclusive).
Try again...
and, force the User make yet another entry.

How should I re-prompt a user for a valid string?

I'm not the best programmer and am quite new to it. I've been trying for hours to get this program correct but I can't seem to come up with a way to make it work out the way I'd like to. Here's what I want to do:
Write a program that prompts a user for their name and then displays "Hello, [Name Here]!"
If the user does not enter anything but pressed Enter anyways, you should re-prompt for the user's name. This flow should look like the following:
Whats is your name?
Please Enter your name:
Please Enter your name: Programming Practice
Hello, Programming Practice!
Here's how I'm thinking of the program before I start writing anything in my IDE:
Ask the user for their name
Give them a chance to enter their name
If their entry is not in name format, give them output saying incorrect format
Give them a chance to enter in proper format
Repeat steps 4 and 5 as many times as it takes for them to enter proper format
Print "Hello, [Name Here]!"
END
Here's what I've got so far:
package lol;
import java.util.Scanner;
public class Whatever {
public static void main(String[] args) {
Scanner sc = new Scanner(System.in);
System.out.printf("What is your name?\n");
String name = sc.nextLine();
if (name != "Programming Practice")
{
System.out.println("Please enter a valid name");
String name2 = sc.nextLine();
System.out.println("Hello, " + name2 );
}
else
{
System.out.println("Hello, " + name );
}
}
}
Right now the output I'm getting regardless of my entries are:
What is your name?
Please enter a valid name
Hello,

You can throw the sc.nextLine() into a while(true) loop and break out when you get a result you deem valid.
String name = "";
while(true) {
name = sc.nextLine();
if (name.equals("Programming Practice")) {
System.out.println("Hello, " + name);
break;
} else {
System.out.println("Please enter in the correct format");
continue;
}
}

A better way of approaching this problem would be by using do-while loop.
do {
// Your processing
} while (checkCondition());
The logic is based on the idea that the user will be prompted to enter details at least once.

java program how add a function that goes back code if player has inputted it wrong

I have a problem with my program is not with the code is how I am going to do it that's the confusing part that I am stuck with. just to let you know I am a basic java coder I do not understand complicated stuff so bear in mind that my code isn't the best.
----------------------------------------------------------- program explaintion-----------------------------------------------------------------
let's get into the point of explaining how it works before I show you my problem, ok when you execute the program it prompts you a sort of like a menu in a video game but it's a text-based, it shows you different options like enter player details, play the math game show score and then quit. enter player details it tells player 1 to enter he/she name and then tells another one to input he/she player name then prompts you back to the menu. play the math game is where a player 1 is asked to input he/she math equation after that player 2 has to solve it if he gets it right he gets 10 points if no the player gets no points at all. then repeats for another player to input he/she math equation then prompts you back to the menu. show scores it shows who got the most scores in the math game it calculates who's got the most if both of them got the same score then means a tie then prompts you back to the menu. and the last thing the quit option when you choose that option it stops the program. if the player chooses a wrong choice he gets an error message and puts you back to the menu
ok here is the first class called menu and other class which is connected with menu called game factions
menu:https://gist.github.com/LOLMEHA/86ff28b038d85030e346785e953644e0
gamefactions:https://gist.github.com/LOLMEHA/f05a51e07c8823a0e65cebbf81cc52ef
so this section of code that I have trouble fingering it out myself
import java.util.*;
public class Gamefunctions // this is a core when player choosess one of these options from the menu
{
String[] player =new String[2];
double scorea = 0; // verribles of all the objects
double scoreb = 0;
int i;
Scanner input = new Scanner(System.in);
double answer = 0;
double numA, numB;
char operator;
char operator2;
boolean quit = false;
double sum1;
double sum2;
public void enterDetails(){ // if player select enter details
for ( i=0;i<2;i++) {// tell's player to input he/she's name and stores them
int c=i;
System.out.println("Welcome to the maths quiz game please input player name "+c++);
player[i] = input.next();
}
}
public void mathGame(){ // if player select enter details
System.out.println("Please enter your equation please "+player[0]+" press enter for each number and mathematical symbol"); // tells the player 1 to input
System.out.println("");
System.out.println("such as for ex input a number or how many you like, then hit enter and input such as /*-+^ hit enter, then input any number one or how many you like ");
String s=input.next();
numA = Double.parseDouble(s); // numa and numb and operator is the aera of player to input he/she equation
operator = input.next().charAt(0);
numB = input.nextDouble();
if () {
if (operator == '+') {// this is if operator is one of these like +-*/^ and then it works out the sum
answer = numA + numB;
}
if (operator == '-') {
answer = numA - numB;
}
if (operator == '*') {
answer = numA * numB;
}
if (operator == '/') {
answer = numA / numB;
}
if (operator == '^') {
answer = Math.pow(numA, numB);
}
} else {
System.out.println("error input like for an example '10' enter '+' enter '10'");
}
System.out.println("");
System.out.println(player[1]+"\t solve the equation"); // tells other player to slove the equation
sum2 = input.nextDouble();
if (sum2 == answer){// checks if the answer from the player is good or not if its good he/she gets 10 points if he/she gets it wrong gets no points and shows the right answer so the player learns from his/she mistakes
scoreb = scoreb + 10.00;
System.out.println("correct you got 10 points to your score");
System.out.println("");
} else{
System.out.println("incorrect you got no points the correct answer was:"+"" + answer);
}
you know when the program ask to player to input his math eqtion and outputs this and continues with the program and waiting for the user to input
public void mathGame(){ // if player select enter details
System.out.println("Please enter your equation please "+player[0]+" press enter for each number and mathematical symbol"); // tells the player 1 to input
System.out.println("");
System.out.println("such as for ex input a number or how many you like, then hit enter and input such as /*-+^ hit enter, then input any number one or how many you like ");
String s=input.next();
numA = Double.parseDouble(s); // numa and numb and operator is the aera of player to input he/she equation
operator = input.next().charAt(0);
numB = input.nextDouble();
let's say that the player inputs like this 10+10 enter but it will not work since they are stored in numA which is an int, I want to make a error message saying that you can not input like this 10+10 you have to input like this 10 enter + enter 10 enter so it will be able to work
if the player inputs it correctly it will continue the program
so if you have any problems with my explaintion of my plroblem pls ask so I can edit it thank you for time :)

Here’s the bit of your code I’m going to be looking at:
String s = input.next();
numA = Double.parseDouble(s);
operator = input.next().charAt(0);
numB = input.nextDouble();
if (/* Some condition */) {
// Calculate answer
} else {
System.out.println("error input like for an example '10' enter '+' enter '10'");
}
First up, a couple nitpicks: Java is not C. You don’t need to declare all your variables at the beginning of your code blocks. numA, numB and operator are never used outside this bit of code, so it makes sense to declare them in here as well.
You’re also using input.next() with Double.parseDouble() once, then input.nextDouble() the next time. Stick to one or the other, it’ll make debugging easier if something doesn’t work properly.
And finally, what happens if someone enters 10 +1 0? The error is silently ignored because the 1 gets picked up as part of the operator string then discarded by charAt(0). A more resilient parsing method here would be to fetch the entire String first, then check for length == 1 before calling charAt(0).
double numA = input.nextDouble();
String operatorString = input.next();
char operator;
if (operatorString.length() == 1) {
operator = operatorString.charAt(0);
} else {
// Handle error
}
double numB = input.nextDouble();
if (/* Some condition */) {
// Calculate answer
} else {
System.out.println("error input like for an example '10' enter '+' enter '10'");
}
Onto your question then: how do we detect an invalid input? Take a look at the documentation for Scanner#nextDouble() (emphasis mine):
public double nextDouble()
Scans the next token of the input as a double. This method will throw InputMismatchException if the next token cannot be translated into a valid double value. If the translation is successful, the scanner advances past the input that matched.
So we know nextDouble() can detect the invalid input for us. It does this in the form of an exception, which we can listen for (or catch) using a try ... catch statement:
try {
double numA = input.nextDouble();
} catch (InputMismatchException e) {
System.err.printf("Invalid input! Expected number, found '%s'.\n", input.next());
}
We could extend this and wrap the entire section of code in a single try ... catch, but then the user would have to start again if they make one mistake. A more user-friendly solution would be this:
double numA;
while (1) {
try {
numA = input.nextDouble();
break;
} catch (InputMismatchException e) {
System.err.printf("Invalid input, try again! Expected number, found '%s'.\n", input.next());
}
}
Note the even if you don’t print it, the call to input.next() is necessary to prevent an infinite loop.
Now we just need to do something similar for operator:
char operator;
while (1) {
String operatorString;
try {
operatorString = input.next();
if (operatorString.length() != 1) {
throw new InputMismatchException();
}
operator = operatorString.charAt(0);
break;
} catch (InputMismatchException e) {
System.err.printf("Invalid input, try again! Expected character, found '%s'.\n", operatorString);
}
}
This seems very similar to the previous snippet for a number - let’s try to refactor some of the common code here into a method:
#FunctionalInterface
public interface ScannerGetter<T> {
T apply() throws InputMismatchException;
}
public <T> T getValueFromScanner(ScannerGetter<T> getter, String type) {
while(1) {
try {
return getter.apply();
} catch (InputMismatchException e) {
System.err.printf("Invalid input, try again! Expected %s.");
}
}
}
There’s a lot going on in these few lines. The first part declares a functional interface - this is a basically a custom type of lambda function. We’ll come back to that in a moment.
The method itself (getValueFromScanner()) is a generic method. It allows us to use the same method with different types in place of the generic parameter (T) without duplicating it.
This is how you’d use the above method to retrieve your three inputs:
double numA = this.<Double>getValueFromScanner(() -> input.nextDouble(), "number");
char operator = this.<Char>getValueFromScanner(() -> {
operatorString = input.next();
if (operatorString.length() != 1) {
throw new InputMismatchException();
}
return operatorString.charAt(0);
}, "operator");
double numB = this.<Double>getValueFromScanner(() -> input.nextDouble(), "number");
// Once your code gets here, all three variables will have valid values.

How to make Java detect keypress in my scenario?

I am writing a program for a project for school. The project requires me to create a record of pets for a pet hospital. I am supposed to create a list of pet owners (maximum of 30 owners) with their first name, last name, e-mail, and phone, and an array pets for each owner (maximum of pets per owner is 5). However, the user can create a list of less than 30 owners. The issue I have is I want to make code where the user can press the escape key to stop the outside while loop that asks them for the owner info when they don't need to add anymore owners. Also, where it says, "System.out.println("Press enter to add another owner or Esc to finish list.");" it is in another loop, so if the user hits an incorrect key it will loop asking them if they want to stop adding tot he list or not (until they hit esc or enter). (Note: the code below is not my whole program, I just needed help with one specific part. I left the if statements blank, because that is where my keypress code will be.).
public class Runner
{
public static void main(String[]args)
{
while(i <= 30)
{
System.out.println(i + ". " + "Enter the owner's first name.");
String first = scan.nextLine();
System.out.println(i + ". " + "Enter the owner's last name.");
String last = scan.nextLine();
System.out.println(i + ". " + "Enter the owner's email address.");
String emailAdd = scan.nextLine();
System.out.println(i + ". " + "Enter the owner's phone number.");
String phone = scan.nextLine();
Owner owner = new Owner(first, last, emailAdd, phone);
int j = 1;
while(j > 0)
{
System.out.println("Press enter to add another owner or Esc to
finish list.");
if ()
{
}
if ()
{
}
j++;
}
list.add(owner.toString());
i+=1;
}
System.out.println(list);
}
}

Since Java is merely using the console, there's no native way to detect a single keypress... any input must end with an "enter" to actually be taken by the program.
As such, to emulate detecting an enter keypress you could do something like:
if (scan.nextLine().isEmpty())
// do stuff
Since we know that a line always means that the enter key was pressed, if the String is empty, it means that only the enter key remains.
For the ESC key, you should use a 3rd party library if is imperative to have it work like that, otherwise, you could have the user actually write "ESC" or some other key word defined by you as input.

I made this program and it prints one extra thing

Program uses a while loop menu in the main to request for the user command:
public static void main(String[] args)throws Exception
{
Boolean meow = true;
while(meow)
{
System.out.println("\n 1. Show all records.\n"
+ " 2. Delete the current record.\n"
+ " 3. Change the first name in the current record.\n"
+ " 4. Change the last name in the current record.\n"
+ " 5. Add a new record.\n"
+ " 6. Change the phone number in the current record.\n"
+ " 7. Add a deposit to the current balance in the current record.\n"
+ " 8. Make a withdrawal from the current record if sufficient funds are available.\n"
+ " 9. Select a record from the record list to become the current record.\n"
+ " 10. Quit.\n");
System.out.println("Enter a command from the list above (q to quit): ");
answer = scan.nextLine();
cmd.command(answer);
if(answer.equalsIgnoreCase("10") || answer.equalsIgnoreCase("q"))
{
meow = false;
}
}
}
If none of the commands you pick are actually commands on the menu then this happens:
else
{
System.out.println("Illegal command");
System.out.println("Enter a command from the list above (q to quit): ");
answer = scan.nextLine();
command(answer);
}
Whenever I add a new person or use any command that requires me to press return to finish entering a value I get the else statement and then the regular command request.
So it looks like:
Enter a command from the list above (q to quit):
Illegal command
Enter a command from the list above (q to quit):
When this happens.
Not gonna post my full code on here, I'm afraid of it cause it's so much. Have the pastebins of them instead.
My full Main Class: http://pastebin.com/rUuKtpXb
My full not Main Class: http://pastebin.com/UE4H76Cd
Anyone know why this is happening?

The problem is that something like Scanner::nextDouble doesn't read the new-line character, so the next Scanner::nextLine returns an empty line.
Replacing all occurrences of Scanner::nextLine with Scanner::next should fix it.
You can also do a Scanner::nextLine after your last non-nextLine next method, but this is a bit messy.
Some other things I'd recommend:
Add scan.useDelimiter("\n"); at the beginning of your program, test with adding spaces to a line and you'll see why this is needed.
Change println to print, so the command can be entered on the same line. i.e.:
Change
System.out.println("Enter a command from the list above (q to quit): ");`
to
System.out.print("Enter a command from the list above (q to quit): ");
Change this:
else
{
System.out.println("Illegal command");
System.out.println("Enter a command from the list above (q to quit): ");
answer = scan.nextLine();
command(answer);
}
to:
else System.out.println("Illegal command");
You would print the menu again, but you would avoid unneeded recursion. It would be easy enough to avoid printing the menu again.
It would be better to check for exit before running command (and then you can remove that check in command).
System.out.println("Enter a command from the list above (q to quit): ");
answer = scan.nextLine();
if (answer.equalsIgnoreCase("10") || answer.equalsIgnoreCase("q"))
meow = false;
else
cmd.command(answer);
Change Boolean to boolean. Boolean is the wrapper class for boolean, which is unneeded in this case.

Maybe its leaving \n in the buffer at the end of the while loop and just before taking input again.
Maybe
while(meow)
{
scan.nextLine();
This can help remove it.