Develop Reference

nextLine() issues with Scanner

I have two do-while loops for making custom input validations. The problem is that it automatically enters in the next do-while loop. I have to put a new nextLine() after I correctly insert the name: name = scanner.nextLine();
I am aware of the "glitch" of nextInt() when the cursor stays there and you have to call nextLine() for it to continue. Source: https://www.geeksforgeeks.org/why-is-scanner-skipping-nextline-after-use-of-other-next-functions/
But this is not the case. I am clearly missing something...
String name = "";
boolean flag_name= false;
do{
System.out.print("Name: ");
if(scanner.hasNextInt()){
System.out.println(scanner.nextInt() + " That's not a valid name...\n");
scanner.nextLine();
}else{
name = scanner.nextLine();
flag_name = true;
}
}while(!flag_name);
int age = 0;
boolean good_age = false;
do {
System.out.print("Age: ");
if (!scanner.hasNextInt()){
System.out.println("That's not a valid age.");
}else if(scanner.nextInt() <= 3 || scanner.nextInt() >= 125) {
System.out.println("You must be over 3yo.");
scanner.nextLine();
}else{
age = Integer.parseInt(scanner.nextLine());
good_age = true;
}
}while (!good_age);
Output:
Name: mark
Age: 'mark' That's not a valid age.
Age:

I finally figured out that I already came with a messed up scanner. On the previous step I had an unattended scanner.next(); which made everything that comes after it behaving erratically.
System.out.print("Insert an option (1,2,3): ");
input = scanner.next();
// changed:
input = scanner.nextLine();
That fixed the scanner for the subsequent steps.
I also applied the logic mentioned by #AppleCiderGuy and fixed my second do-while loop.
Final code:
String name = "";
boolean flag_name= false;
do{
System.out.print("Name: ");
if(scanner.hasNextInt()){
System.out.println("That's not a valid name...");
scanner.nextLine();
}else{
name = scanner.nextLine();
flag_name = true;
}
}while(!flag_name);
int age = 0;
boolean good_age = false;
do {
System.out.print("Age: ");
if (!scanner.hasNextInt()) {
System.out.println("That's not a valid age.");
scanner.nextLine();
}else{
age = Integer.parseInt(scanner.nextLine());
if(age <= 3 || age >= 125) {
System.out.println("You must be over 3yo.");
}else{
good_age = true;
}
}
}while (!good_age);
Lessons learned:
You will always need a scanner.nextLine(); for cleaning the scanner when incorrect inputs are reached. Otherwise the incorrect input will stay in the scanner and mess up the subsequent code.
When you use integers, even if correctly assigned to variables, you'll also need to clean the scanner. Is easier for me to use Integer.parseInt(scanner.nextLine()); instead.
Always try to evaluate variables instead of scanner.next* methods. Because the scanner will be waiting more inputs.
Thanks.

What is wrong with this java while loop?

New to Java and learning how to use While loops and random generator. This prints a multiplication question. Every time the user answers a question wrong, it should print the same question. Instead, it exits the program. What should I do?
while (true) {
Random multiply = new Random();
int num1 = multiply.nextInt(15);
int num2 = multiply.nextInt(15);
int output = num1 * num2;
System.out.println("What is the answer to " + num1 + " * " + num2);
Scanner input = new Scanner(System.in);
int answer = input.nextInt();
if (answer == output) {
if (answer != -1)
System.out.println("Very good!");
} else {
System.out.println("That is incorrect, please try again.");
}
}

If you want to repeat the same question when the user gets the answer wrong, you should use another while inside your main loop.
This inner loop continues to ask as long as you give a wrong answer.
I also replaced nextInt with nextLine, which reads in a whole line of text. This consumes the "Enter" key and is a safer approach at reading from the console. Since the result is now a String you need to use Integer.parseInt to convert it to an int. This throws an exception if you enter anything but a whole number so I wrapped it into a try-catch block.
If you want, you can add an additional check for validating user input. So in case the user wants to stop playing they only need to input "exit" and the whole outer loop will exit.
boolean running = true; // This flag tracks if the program should be running.
while (running) {
Random multiply = new Random();
int num1 = multiply.nextInt(15);
int num2 = multiply.nextInt(15);
int output = num1 * num2;
boolean isCorrect = false; // This flag tracks, if the answer is correct
while (!isCorrect) {
System.out.println("What is the answer to " + num1 + " * " + num2);
Scanner input = new Scanner(System.in);
try {
String userInput = input.nextLine(); // Better use nextLine to consume the "Enter" key.
// If the user wants to stop
if (userInput.equals("exit")) {
running = false; // Don't run program any more
break;
}
int answer = Integer.parseInt(userInput);
if (answer == output) {
if (answer != -1) {
System.out.println("Very good!");
isCorrect = true; // Set the flag to true, to break out of the inner loop
}
} else {
System.out.println("That is incorrect, please try again.");
}
}
catch(NumberFormatException e) {
System.out.println("Please enter only whole numbers");
}
}
}

Avoid while true. Declare a variable to true, pass the variable to the condición loop and set it to false when the answer is incorrect. You can use break too, but is easier to read the code when you use a exit condition in the while. Also read more about loops https://docs.oracle.com/javase/tutorial/java/nutsandbolts/while.html

java : how can i use if correctly

I recently started java programming
but I have a problem
i want to write a program. I have a password, I ask the user of the program to enter the password
I want: if the person entered a string, I tell him that please don't enter string
and if the password was right and the type of the password that he entered(int) was right, I tell him OK.
in the test of the program, my problem is that when I entered a wrong password and expect that the program tell me that the pass is wrong, the program just tell me nothing !!
here is my code :
int pass = 123 ;
Scanner password = new Scanner(System.in);
System.out.println("Please Enter Your Password : ");
if (password.hasNextInt() && password.nextInt()==pass)
{
System.out.println("ok");
}
else if (password.hasNextInt())
{
System.out.println("wrong pass");
}
else
{
System.out.println("wrong type");
}

You are using hasNextInt() From Java docs.
Returns true if the next token in this scanner's input can be
interpreted as an int value
So you are asking twice for the input.
Example
Input:
1234 (first Input)
1234 (Then hasNextInt() is asking for input again)
OutPut :
wrong pass
So I made this simple snippet for you can use
Scanner password = new Scanner(System.in);
System.out.println("Please Enter Your Password : ");
int pass = 123;
try {
int myInput = password.nextInt();
if (myInput == pass) {
System.out.println("ok");
}else{
System.out.println("wrong pass");
}
}catch (java.util.InputMismatchException e) {
System.out.println("wrong type");
}

The problem is that Scanner methods like nextInt() consume input that's then no longer available to later Scanner calls.
int pass = 123 ;
Scanner password = new Scanner(System.in);
System.out.println("Please Enter Your Password : ");
if (password.hasNextInt() && password.nextInt()==pass) // line A
{
System.out.println("ok");
}
else if (password.hasNextInt()) // line B
{
System.out.println("wrong pass");
}
else
{
System.out.println("wrong type");
}
So in case of entering a wrong password, e.g. 4321, what happens?
Line A checks password.hasNextInt() as the first half of your condition. The Scanner doesn't know that right now and waits for your console input. You enter 4321, and now the Scanner can check whether that's a valid number (and it does so without consuming the 4321, so that it's still available). It is, so the program continues to the next part of the condition (side remark: were it abc, that first part would be false, and Java would already know that the combined password.hasNextInt() && password.nextInt()==pass condition would be false, without a need to go into the second half, thus not consuming the entry).
Line A now checks the second half password.nextInt()==pass. This calls nextInt(), returning the integer 4321 and consuming the input. Comparing this against your number 123 gives false, so the condition doesn't match. That's what you want so far.
Now in line B you want to check for the case of a number not being 123. But your condition password.hasNextInt() no longer sees the 4321 we entered, as that has been consumed in line A. So it waits for the next input. That's the problem, you're still calling hasNextInt() after consuming the input with nextInt().
You can change your program like this:
int pass = 123 ;
Scanner password = new Scanner(System.in);
System.out.println("Please Enter Your Password : ");
if (password.hasNextInt()) {
if (password.nextInt()==pass) {
System.out.println("ok");
} else {
System.out.println("wrong pass");
}
} else {
pass.next(); // consume the invalid entry
System.out.println("wrong type");
}
[ I reformatted the code snippet in a more Java-typical style, doesn't change the functionality of course, but looks more familiar to me. ]
Of course, Gatusko's exception-based approach works as well, and personally I'd do it his way, but maybe you don't feel comfortable with exceptions right now, so I stayed as close to your approach as possible.

You can use the following piece of code.
public static void main(String[] args){
int pass = 123 ;
Scanner password = new Scanner(System.in);
System.out.println("Please Enter Your Password : ");
if (password.hasNextInt())
{
if(password.nextInt()==pass) {
System.out.println("ok");
}
else {
System.out.println("Wrong password");
}
}
else
{
System.out.println("wrong type");
}
}

What about a while?
int MAX_TRIES = 3
int currentTries = 0;
while (password.hasNextInt() && currentTries < MAX_TRIES) {
if (password.nextInt()==pass) {
// OK!
} else {
// Wrong!
}
currentTries++;
}
if (currentTries == MAX_TRIES) {
// You tried too much
} else {
// Password was a string
}

Try this code, if the input is not an integer then it will throw NumberFormatException, which is caught and displayed.
public static void main(String[] args){
int pass = 123 ;
Scanner password = new Scanner(System.in);
System.out.println("Please Enter Your Password : ");
String enteredPassword ="";
if(password.hasNext() && (enteredPassword = password.next()) !=null){
try{
if(Integer.parseInt(enteredPassword) == pass){
System.out.println("ok");
}else{
System.out.println("wrong pass");
}
}catch (NumberFormatException nfe){
System.out.println("wrong type");
}
}
}

Scanner loop doesn't cycle when hitting enter twice

boolean loop = false;
double numberOfStudents;
System.out.print("Enter a number: ");
if ((scnr.nextLine().trim().isEmpty()) ) {
loop = true;
}
while (loop) {
System.out.println("Enter a number");
if (scnr.hasNextDouble() ){
System.out.println("Loop has stopped");
numberOfStudents = scnr.nextDouble();
loop = false;
}
}
System.out.println("You're outside the loop!");
I'm trying to get the program to say "Enter a number" until the user has entered an actual number (no white spaces or letters or signs). When the user has entered a number, it sets numberOfStudents equal to that number and breaks out of the loop.
But if you hit enter twice, it doesn't iterate. It only displays "Enter a number" once.
What is wrong with the loop logic? Why isn't it looping until valid input is taken?

For the actual answer to your question of "Why doesn't 'Enter a number' display more than once?" see Tom's comment (update: Tom's answer).
I've rewritten your loop in a way which preserves your code, but also makes it a little easier to handle format exceptions (though at the risk of silently swallowing an exception -- should be acceptable for this use case).
Can be up to you to use this design, here is an SO post on why empty catch blocks can be a bad practice.
public static void main(String args[])
{
boolean loop = true;
double numberOfStudents;
Scanner scnr = new Scanner(System.in);
while(loop){
System.out.print("Enter a number: ");
String input = scnr.nextLine();
try{
numberOfStudents = Double.parseDouble(input);
loop = false;
}catch(NumberFormatException e){
}
}
System.out.println("You're outside the loop!");
}
Output:
Enter a number:
Enter a number:
Enter a number:
Enter a number: 50
You're outside the loop!

First of all: Since you're reading from System.in a call to the input stream will block until the user entered a valid token.
So let's check first scan using your scnr variable:
scnr.nextLine()
nextLine() reads everything til the next line delimiter. So if you just press return, then it will successfully read it and will perform the next stuff.
The next call is:
scnr.hasNextDouble()
This call expects a "real" token and ignores white spaces, except as a delimiter between tokens. So if you just press return again it doesn't actually read that input. So it still waits for more (for the first token). That is why it stucks in your loop and you won't get another "Enter a number" output.
You can fix that by either enter a real token, like a number, or by changing the loop like trobbins said.
I hope you now understand your program flow a bit more :).

While trobbins code basically solves your problem, it's bad practice to use exceptions for flow control.
I used a small regexp to check if the value is a number. But this example is not complete, it will still crash it the user enters for example two decimal points. So you would need to create a proper number check or just use integers where the check is much easier.
Someone in the comments pointed out that people may want to enter scientific notation like 5e10, so this would also be another case to check for. If this is just some code you need as a proof of concept or something quick and dirty, you can go with the exception handling method but in production code you should avoid using exceptions this way.
double numberOfStudents;
Scanner scnr = new Scanner(System.in);
while(true) {
System.out.print("Enter a number: ");
String input = scnr.nextLine().trim();
if(input.matches("^[0-9\\.]{1,}$")) {
System.out.println("Loop has stopped");
numberOfStudents = Double.parseDouble(input);
break;
}
}
System.out.println("You're outside the loop!");

The following code should help you:
double numberOfStudents = 0;
Scanner scnr = new Scanner(System.in);
boolean readValue = false; //Check if the valid input is received
boolean shouldAskForNumber = true; //Need to ask for number again? Case for Enter
do {
if (shouldAskForNumber) {
System.out.print("Enter a number:");
shouldAskForNumber = false;
}
if (scnr.hasNextDouble()) {
numberOfStudents = scnr.nextDouble();
readValue = true;
} else {
String token = scnr.next();
if (!"".equals(token.trim())) { //Check for Enter or space
shouldAskForNumber = true;
}
}
} while (!readValue);
System.out.printf("Value read is %.0f\n", numberOfStudents);
System.out.println("You're outside the loop!");
Update
Understood the following statement in question different way:
But if you hit enter twice, it doesn't loop back. It only displays
"Enter a number" once.
The code is set to print "Enter a number" only once if the user hits RETURN/ENTER or enters space character. You may remove the special check and use the code if needed.

import java.util.Scanner;
public class Testing {
public static boolean checkInt(String s)
{
try
{
Integer.parseInt(s);
return true;
} catch (NumberFormatException ex)
{
return false;
}
}
public static void main(String[] args) {
boolean loop = false;
double numberOfStudents;
Scanner scnr = new Scanner(System.in);
String input = "";
while (!(checkInt(input))) {
System.out.println("Enter a number");
input = scnr.nextLine();
}
numberOfStudents = Integer.parseInt(input);
System.out.println("Number of students: " + numberOfStudents );
}
}
//this code is working fine, if you want you check it out.
//In your code your taking another input if the first is an int/double; if the first input is not a number then you have mentioned to take input again..

Use a debugger to see what the code is actually doing. Here's a guide on debugging in Eclipse. After you have finished debugging your code, you will probably know what the problem is.

Below code will help you
boolean loop = true;
double numberOfStudents;
Scanner scnr = new Scanner(System.in);
System.out.print("Enter a number: ");
String input = scnr.nextLine();
while(!scnr.hasNextDouble()){
System.out.print("Enter a number: ");
try{
numberOfStudents = Double.parseDouble(input);
break;
}catch(NumberFormatException e){
}
input = scnr.nextLine();
}
System.out.println("You're outside the loop!");

The following code is working,
boolean loop = true;
double numberOfStudents;
Scanner scnr=new Scanner(System.in);
while(loop) {
System.out.println("Enter a number");
if ((scnr.nextLine().trim().isEmpty()) ) {
loop = true;
}
if (scnr.hasNextDouble() ){
System.out.println("Loop has stopped");
numberOfStudents = scnr.nextDouble();
loop = false;
}
}
System.out.println("You're outside the loop!");
The output is,
run:
Enter a number
hj
po
Enter a number
lhf
Enter a number
o
Enter a number
p
Enter a number
a
Enter a number
34
Loop has stopped
You're outside the loop!

You have to scan the next line if you want to get more values form the scanner again. The code should be like:
while (loop) {
System.out.println("Enter a number");
if(!(scnr.nextLine().trim().isEmpty())){
if (scnr.hasNextDouble() ){
System.out.println("Loop has stopped");
numberOfStudents = scnr.nextDouble();
loop = false;
}
}
}

Java input keeps being empty?

This is a simple question selection, and then answer program:
import java.util.Scanner;
public class Mains {
static Scanner console = new Scanner(System.in);
static Tof tof = new Tof();
static int Ievel = 0;
static int Input = 0;
static boolean GAME = true;
static boolean AT_START = true;
static boolean IN_QUESTION = false;
public static void main (String[] args) {
while (GAME) {
String InputS = "";
if (AT_START) {
System.out.println("Welcome to the game! Please select a number from 1 to 10.");
AT_START = false;
}
if (!IN_QUESTION)
Input = console.nextInt();
if (Input == -1) {
GAME = false;
console.close();
} else {
String question = tof.getQuestion(Input);
String answer = tof.getAnswer(Input);
System.out.println(question);
IN_QUESTION = true;
while (IN_QUESTION) {
InputS = console.nextLine();
if (InputS != console.nextLine()) {
if (InputS.equals(answer)) {
System.out.println("Correct!");
} else {
System.out.println("Incorrect. " + InputS + " " + answer);
}
}
}
}
}
}
}
Problem:
When entering the IN_QUESTION loop, and writing a answer, it will always be incorrect.
That's because the InputS variable is ALWAYS empty, no matter what, while it has console.nextLine() set on it.
Why is it empty? How do I fix this?
In-case you need the other class Tof: http://pastebin.com/Fn5HEpL2

nextInt doesn't get the line terminator after the integer and you're reading from the console twice (the second time being in the if-statement).
So if you enter:
123
apple
The following happens:
Input gets assigned a value of 123
InputS gets assigned an empty string
InputS gets compared against apple and it is not equal (from InputS != console.nextLine() - I'm not sure why it's there)
You can fix it by:
Putting a console.nextLine(); after console.nextInt();
OR
Use Input = Integer.parseInt(console.nextLine()) instead of nextInt
Removing this - if (InputS != console.nextLine())

You're reading from the console twice. This should work:
while (IN_QUESTION) {
InputS = console.nextLine();
if (InputS.equals(answer)) {
System.out.println("Correct!");
} else {
System.out.println("Incorrect. " + InputS + " " + answer);
}
}

The problem is that the new line character was not read by the nextInt() method so it remain in the scanner buffer and when the next time you called nextLine() that character was printed first.
This is how to fix the issue:
//empty the newline character from the scanner
console.nextLine();
while (IN_QUESTION) {
InputS= console.nextLine();
if (InputS.equals(answer)) {
System.out.println("Correct!");
} else {
System.out.println("Incorrect. " + InputS + " " + answer);
}
}

You call console.nextLine twice. This means that you read a line that you'll check, and another you won't. This is probably not what you are after. Also note that your initial call of nextInt will not consume the newline you pressed after entering the number. You need a nextLine right after that, but before the main loop.
Some general remarks:
uppercase names are only for constants, so your variables should be lowercase;
you should really be using local variables instead of static ones. Right now this isn't hurting you, but it soon could.