This question already has answers here:
Scanner is skipping nextLine() after using next() or nextFoo()?
(24 answers)
Closed 8 years ago.
I was writing my first, well second if you count hello world, program and ran into a small issue.
My code:
import java.util.*;
class test {
public static void main(String[] args) {
Scanner scan = new Scanner(System.in);
System.out.print("What is your name: ");
String name = scan.nextLine();
System.out.print("What is your favorite number: ");
int favoriteNumber = scan.nextInt();
System.out.print("What is your favorite game: ");
String game = scan.nextLine();
}
}
So it scans in name and favoriteNumber but the program ends before asking the user to enter a favorite game. Essentially the user can only enter text twice instead of three times like I would like.
Just a guess: scan.nextInt() does not consume the trailing newline character after the user presses enter. The subsequent call to scan.nextLine() then finds a newline character waiting.
Try switching out int favoriteNumber = scan.nextInt(); for String favoriteNumber = scan.nextLine(); to see if that fixes the issue. If it does, my hypothesis is correct.
If that is the case, then you should probably use Integer.parseInt to convert this string into an integer. The problem here is effectively that you really want to collect 3 lines of input, where a line is defined as "any sequence of characters ending with a newline". Your program is currently written to request a line, an int, then a line, rather than "3 lines of input, one of which contains an int".
an exception must be thrown in here scan.nextInt();
Check whether you entered a proper int
Related
This question already has answers here:
Scanner is skipping nextLine() after using next() or nextFoo()?
(24 answers)
Closed 1 year ago.
I'm new to java and I wrote this code
import java.util.Scanner;
public class GamingJava {
public static void main(String[] args) {
Scanner input = new Scanner(System.in);
String Name;
int password;
String yEs;
System.out.print("hello sir what is your name? ");
Name = input.nextLine();
System.out.print("what is your password? ");
password = input.nextInt();
System.out.println("your name was "+Name+" and your password was "+password);
System.out.print("are you sure? ");
yEs = input.nextLine();
System.out.println(yEs);
}
}
It only ask the name and the password, but Java doesn't ask the last one how did that happen?
It asks for the input and immediately takes it as an empty string i.e. "".
Root Cause : The issue is because of the nextLine() method. Since while providing an input, user has to press the Enter key, the cursor/prompt on the console moves to next line. This line is taken as an empty line by the nextLine() method.
Solution : You must use the next() method as it looks out for the presence of space character for taking the input.
FYR the following line of code
yEs = input.nextLine();
should be changed to
yEs = input.next();
This question already has answers here:
Scanner is skipping nextLine() after using next() or nextFoo()?
(24 answers)
Closed 7 years ago.
I know a lot of people have asked this but everyones answer is to use scanner.nextLine() however this does not work for me. I have this code:
if (option == 1) {
System.out.println("What is the name of the goal the task is under?: ");
String goalName = scanner.next();
}
When I use .next() I can only get one word, however when I use .nextLine() the command line does not allow for the user to input anything so it asks "What is the name of the goal the task is under?:" and then continues on the program without letting input occur.
The following test code works:
package asdfsadf;
import java.util.Scanner;
public class Asdfsadf {
public static void main(String[] args) {
Scanner scanner = new Scanner(System.in);
System.out.println("What is the name of the goal the task is under?: ");
String goalName = scanner.nextLine();
System.out.println("You entered: ");
System.out.print(goalName);
}
}
So the only problem would be if you used scanner before this say you had scanner.nextInt() and that would leave an empty "end of the line" in the scanner. So the nextLine would take the end of the line from the previous call. To fix this, just create a new scanner object.
This question already has answers here:
Scanner is skipping nextLine() after using next() or nextFoo()?
(24 answers)
Closed 8 years ago.
I am learning Java and I was trying an input program. When I tried to input an integer and string using instance to Scanner class , there is an error by which I can't input string. When I input string first and int after, it works fine. When I use a different object to Scanner class it also works fine. But what's the problem in this method when I try to input int first and string next using same instance to Scanner class?
import java.util.Scanner;
public class Program {
public static void main(String[] args) {
Scanner input=new Scanner(System.in);
//Scanner input2=new Scanner(System.in);
System.out.println("Enter your number :" );
int ip = input.nextInt();
System.out.println("Your Number is : " + ip);
System.out.println("Enter Your Name : ");
String name= input.nextLine();
System.out.println("Your Next is : " + name);
}
}
nextInt() doesn't wait for the end of the line - it waits for the end of the token, which is any whitespace, by default. So for example, if you type in "27 Jon" on the first line with your current code, you'll get a value of 27 for ip and Jon for name.
If you actually want to consumer a complete line, you might be best off calling input.nextLine() for the number input as well, and then use Integer.parseInt to parse the line. Aside from anything else, that represents what you actually want to do - enter two lines of text, and parse the first as a number.
Personally I'm not a big fan of Scanner - it has a lot of gotchas like this. I'm sure it's fine when it's being used in exactly the way the designers intended, but it's not always easy to tell what that is.
If you call input.nextInt(); the scanner reads the number from the input, but leaves the line separator there. That means, if you call input.nextLine(); next, it reads everything till the next line separator. And this is in this case only the line separator itself.
You can fix that in two ways.
Way 1:
int ip = Integer.parseInt(input.nextLine());
// output
String name= input.nextLine();
Ways 2:
int ip = input.nextInt();
// output
input.nextLine();
String name= input.nextLine();
This one working, Anyway if you want to save IP address it must be String.
import java.util.Scanner;
public class Program {
public static void main(String[] args) {
Scanner input = new Scanner(System.in);
//Scanner input2=new Scanner(System.in);
System.out.println("Enter your number :");
int ip = input.nextInt();
System.out.println("Your Number is : " + ip);
System.out.println("Enter Your Name : ");
input.nextLine();
String name = input.nextLine();
System.out.println("Your Next is : " + name);
}
}
This question already has answers here:
Scanner is skipping nextLine() after using next() or nextFoo()?
(25 answers)
Closed 8 years ago.
I'm trying to make a "Bank Account" and that requires the user to input the name, password, balance, and interest of that bank account. I use a sentinel controlledwhile loop to achieve this, but my sentinel does not work for some reason. Here's part of my code, if you could help me that would be great:
System.out.println("New Account: Enter name, type \"quit\" to exit. ");
String name = scan.nextLine();
while (!name.equals("quit")) {
System.out.println("Enter password: ");
String pass = scan.nextLine();
System.out.println("Enter balance (in pennies): ");
double bal = scan.nextDouble();
System.out.println("Enter interest: ");
double inter = scan.nextDouble();
BankAccount2 bankacc = new BankAccount2(name, pass, bal, inter);
accounts.add(bankacc);
System.out.println("Successful account creation! New Account: Enter name, type \"quit\" to exit. ");
name = scan.nextLine();
}
To clarify, the problem is that the first time when I enter "quit", before the loop starts, it works well and the loop does not begin. However, when I input a bank account, and it goes to "Enter name" all over again, when I type quit the loop does not stop.
I believe that the nextDouble() method does not read the next new line character. Check the documentation about this.
You might try calling nextLine() after calling nextDouble().
You can check the source of Scanner here. If you are stuck because of crappy javadoc or undocumented side effects I think it is always best if you check the code.
It seems that it is using a regex matcher which only matches a double pattern.
This question already has answers here:
Scanner is skipping nextLine() after using next() or nextFoo()?
(25 answers)
Closed 8 years ago.
Consider the following snippet:
System.out.print("input iteration cycles");
Scanner reader=new Scanner(System.in);
int iteration = reader.nextInt();
System.out.print("input choice: Var or par");
String choice=reader.nextLine();
System.out.print(choice);
I was hoping to get one line where it prints the first input and the next print out to only print the next input which is either var or par, however it seems that the second print prints out the second input in addition to the first input. Any idea why?
nextInt() does not advance to the next line. So you must make an explicit call to nextLine() immediately after it. Otherwise, choice will be given that newline. Just add this line to your code:
System.out.print("input iteration cycles");
Scanner reader=new Scanner(System.in);
int iteration = reader.nextInt();
reader.nextLine(); //<==ADD
System.out.print("input choice: Var or par");
String choice=reader.nextLine();
System.out.print(choice);