This question already has answers here:
Scanner is skipping nextLine() after using next() or nextFoo()?
(24 answers)
Closed 1 year ago.
Hey there I was trying to make a phone contact project and I started to create a addContact() void that takes 2 inputs name and phone number in an infinite while loop. My problem is it doesn't take the phone number input after the name input. Can you please help me? Here is my code below
import java.util.*;
public class Problem {
private final Scanner scanner = new Scanner(System.in);
private final ArrayList<String> contactNames = new ArrayList<>();
private final ArrayList<Integer> contactNumbers = new ArrayList<>();
private final HashMap<String, Integer> contactMap = new HashMap<>();
public void addContact() {
while (true) {
System.out.println("Enter a phone number");
int contactNumber = scanner.nextInt();
if (contactNumbers.contains(contactNumber)) {
System.out.println("This number is already given in your contacts");
break;
}
System.out.println();
System.out.println("Enter your contacts name");
String name = scanner.nextLine();
name = name.toLowerCase();
if(contactNames.contains(name)){
System.out.println("This name is already given in your contacts");
break;
}
contactNumbers.add(contactNumber);
contactNames.add(name);
contactMap.put(name, contactNumber);
System.out.println("Person is successfully added to your contacts, here is your all contacts below");
System.out.println(contactMap);
System.out.println("Do you want to quit? -> 1=yes, 2=no");
int quit = scanner.nextInt();
if(quit==1)
break;
}
}
public static void main (String[] args){
Problem a = new Problem();
a.addContact();
}
}
Solution
scanner.nextLine(); //add this line
System.out.println();
System.out.println("Enter your contacts name");
This is to get rid of the newline created once the user has entered their phone number and pressed enter.
More info from here: Scanner is skipping nextLine() after using next() or nextFoo()?
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This question already has answers here:
Scanner is skipping nextLine() after using next() or nextFoo()?
(24 answers)
Closed 2 years ago.
import java.util.Scanner;
public class NoteIt {
public static void main(String[]args) {
Scanner s = new Scanner(System.in);
int Answer;
int i=2;
System.out.print("\nPlease Enter your Name: ");
String Name = s.nextLine();
System.out.println("Welcome to Note-It "+Name+", We hope you'll enjoy our application. ");
String[][] Main = new String[2][2];
Main[0][0]="Create new Note";
Main[1][0]="View My Notes";
System.out.println("\nPlease select what to do: \n");
for(int n=0; n<2; n++){
System.out.println((n+1)+") "+Main[n][0]);
}
System.out.print("\nPlease enter your response: ");
Answer = s.nextInt();
if(Answer == 1){
i++;
Main = new String[i][2];
System.out.print("\nTitle: ");
Main[i-1][0]=s.nextLine();
System.out.print("\nBody: ");
Main[i-1][1]=s.nextLine();
}
}
}
I don't know why it is not asking for Title?
According to #Rohit Jain, That's because the Scanner.nextInt method does not read the newline character in your input created by hitting "Enter," and so the call to Scanner.nextLine returns after reading that newline.
Description here: check this question too
Try this it may help...
Scanner s = new Scanner(System.in);
int Answer;
int i=2;
System.out.print("\nPlease Enter your Name: ");
String Name = s.nextLine();
System.out.println("Welcome to Note-It "+Name+", We hope you'll enjoy our application. ");
String[][] Main = new String[2][2];
Main[0][0]="Create new Note";
Main[1][0]="View My Notes";
System.out.println("\nPlease select what to do: \n");
for(int n=0; n<2; n++){
System.out.println((n+1)+") "+Main[n][0]);
}
System.out.print("\nPlease enter your response: ");
Answer = s.nextInt();
if(Answer == 1){
i++;
Main = new String[i][2];
System.out.print("\nTitle: ");
Main[i-1][0]=s.next();
System.out.print("\nBody: ");
Main[i-1][1]=s.next();
}
This question already has answers here:
Scanner is skipping nextLine() after using next() or nextFoo()?
(24 answers)
Closed 3 years ago.
This is a simple test I made for my user input of my program but it skips over the second user input part. I am using sc.nextLine()
public static void main(String[] args)
{
Scanner sc = new Scanner(System.in);
int option;
String name;
System.out.println("1 or 2");
option = sc.nextInt();
if(option == 1)
{
System.out.println("Please enter a name");
name = sc.nextLine();
System.out.println(name);
}
}
The problem is that the enter ("\n") from the first user input isn't a number and is used for the second.
The quick and dirty option to fix this is to add a "sc.nextLine ();" after "nextInt()" line.
Like:
public static void main(String[] args)
{
Scanner sc = new Scanner(System.in);
int option;
String name;
System.out.println("1 or 2");
option = sc.nextInt();
sc.nextLine();
if(option == 1)
{
System.out.println("Please enter a name");
name = sc.nextLine();
System.out.println(name);
}
}
Use sc.nextLine() after sc.nextInt() in your code.
This question already has answers here:
Scanner is skipping nextLine() after using next() or nextFoo()?
(24 answers)
Closed 4 years ago.
I was trying to take input for the number of names to be stored in an array from user and then using that i was taking names from the user ,first i tried to take names from the user using next() method and all the things were fine but when i tried to take input using nextLine() method the output was as shown below
package learningJava;
import java.util.Scanner;
public class practice
{
public static void main(String[] args)
{
int n;
Scanner obj = new Scanner(System.in);
System.out.println("Enter the number of names you are gonna enter");
n = obj.nextInt();
String names[] = new String[n];
for(int i=0;i<n;i++)
{
System.out.println("Enter the name of friend "+(i+1));
names[i]=obj.nextLine();
}
obj.close();
System.out.println("Names of your friends are");
for(int i=0;i<names.length;i++)
{
System.out.println(names[i]);
}
}
}
Output for the nextLine() method
Enter the number of names you are gonna enter
5
Enter the name of friend 1
Enter the name of friend 2
It is not prompting me to enter the name of friend 1 and directly skipping it and coming to the friend 2 line.
I am beginner in Java , i know the basic difference in next and nextLine() that next() doesn't take input after a space but nextLine() takes complete input , So what is happening here ??
just in for loop, just change "println" to "print" because nextLine() consumes new line character.
import java.util.Scanner;
public class practice
{
public static void main(String[] args)
{
int n;
Scanner obj = new Scanner(System.in);
System.out.println("Enter the number of names you are gonna enter");
n = obj.nextInt();
String names[] = new String[n];
for(int i=0;i<n;i++)
{
System.out.print("Enter the name of friend "+(i+1));
names[i]=obj.nextLine();
}
obj.close();
System.out.println("Names of your friends are");
for(int i=0;i<names.length;i++)
{
System.out.println(names[i]);
}
}
}
check this answer: Java String Scanner input does not wait for info, moves directly to next statement. How to wait for info?
This question already has answers here:
Scanner is skipping nextLine() after using next() or nextFoo()?
(24 answers)
Closed 7 years ago.
Everything is working fine in this code sample that i did, but everytime it calls the do-while loop on the output screen, it skips the "First name" input, and goes straight to "How old are you?" input. why is that happening and how can i fix it? I want to be able to repeat the whole thing without skipping "First name" when i pick 'y' to perform the loop.
public class WeightGoalTest
{
public static void main(String[]args)
{
doWhile person = new doWhile();
person.userInput();
}
}
import java.util.*;
public class doWhile
{
Scanner keyboard = new Scanner(System.in);
private String inputFn = "";
private int inputAge = 0;
private int inputHeight = 0;
private double inputWeight = 0.0;
private int inputCalculation = 0;
private int inputGender = 0;
private char loop;
public void userInput()
{
do
{
System.out.println("First Name: ");
inputFn = keyboard.nextLine();
System.out.println("How old are you?");
inputAge = keyboard.nextInt();
System.out.println("Gender 1 - Male: ");
System.out.println(" 2 - Female: ");
inputGender = keyboard.nextInt();
System.out.println("What is your Height in inches? ");
inputHeight = keyboard.nextInt();
System.out.println("Current Weight in pounds: ");
inputWeight = keyboard.nextDouble();
System.out.println("\nDo you want to use the calculator again? (y/n)");
loop = keyboard.next().charAt(0);
}while(loop == 'y');
}
}
Please note:
String java.util.Scanner.next() - Returns:the next token
String java.util.Scanner.nextLine() - Returns:the line that was skipped
Change your code [do while initial lines] as below:
System.out.println("First Name: ");
inputFn = keyboard.next();
use keyboard.next() instead of .nextLine(). For further informations have a look in the Java API concerning the Scanner class. nextLine() will
Advance this scanner past the current line and return the input that was skipped."
This question already has answers here:
Scanner is skipping nextLine() after using next() or nextFoo()?
(24 answers)
Why can't I enter a string in Scanner(System.in), when calling nextLine()-method?
(13 answers)
Closed 9 years ago.
I used Java and I tried to write program that reads a number of patients and then let the employee enter their names and ages.
and I want the program reads full name such as (Maha Saeed) so I wrote the code like this but I don't know why it does not work
name = scan.nextLine();
and this is the full code
import java.util.*;
public class Answer1
{
static Scanner scan = new Scanner (System.in);
public static void main (String[] args)
{
int age ;
int PatientNumber ;
int PatientNO;
String name ;
System.out.print("Enter number of patients :");
PatientNumber = scan.nextInt();
PatientNO = 1;
while ( PatientNO <= PatientNumber)
{
System.out.println("Patient #" +PatientNO);
System.out.print("Enter patient's Name: ");
name = scan.nextLine(); //<- here is the problem if I write scan.next it works but it reads only the first name
System.out.print("Enter patient's Age: ");
age= scan.nextInt();
PatientNO = PatientNO + 1;
}
}
}
thanks all
See Why can't I enter a string in Scanner(System.in), when calling nextLine()-method? and Scanner is skipping nextLine() after using next(), nextInt() or other nextFoo() methods
Simple solution, you can consume the \n character:
scan.nextLine();
name = scan.nextLine();