In Java Program Scanner is Working only once whereas i/ve used it twice

Question

import java.util.Scanner;

class Student1{
    int a;
    String b;
    void get(int c, String d){
        a = c;
        b = d;
    }
    void show(){
        System.out.print("Id is " + a +"\n");
        System.out.print("Name is " + b);
    }
    public static void main(String args[]){
        Scanner one = new Scanner(System.in);
        System.out.print("Enter Id");
        int e = one.nextInt();
        System.out.print("Enter Your Name");
        String f = one.nextLine();
        Student1 s = new Student1();
        s.get(e ,f);
        s.show();
    }
}

when the program is executed it only asks for Id after that it shows result it never asks for Name

int e = one.nextInt(); just replace your this line with this one int e = Integer.parseInt(one.nextLine()); — Mohammad Ashfaq, May 29 '15 at 10:13

score 2 · Accepted Answer · edited May 23 '17 at 11:51

2

it gets complicated when you use nextInt() and nextLine() together. you can try this version which only uses nextLine();

public static void main(String args[]){
    Scanner one = new Scanner(System.in);
    System.out.print("Enter Id");
    String number =one.nextLine();
    int e = Integer.parseInt(number);
    System.out.print("Enter Your Name");
    String f = one.nextLine();
    Student1 s = new Student1();
    s.get(e ,f);
    s.show();
}

why nextInt() and nextline() produce some problems:

Scanner is skipping nextLine() after using next(), nextInt() or other nextFoo() methods

edited May 23 '17 at 11:51

Community

1
1

answered May 29 '15 at 10:08

nafas

5,283
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why does it get complicated? – user4641842 May 29 '15 at 10:40
@user4641842 I've added a link to the answer regarding your question – nafas May 29 '15 at 10:49

score 1 · Answer 2 · edited May 29 '15 at 10:15

1

You must be entering a newline after integer. For that that blank string is acting as input for one.nextLine(). To avoid that one way is to add one more one.nextLine() statement before String f = one.nextLine();

edited May 29 '15 at 10:15

Soumitri Pattnaik

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answered May 29 '15 at 10:10

Masud Parves Bhuiyan

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score 0 · Answer 3 · answered May 29 '15 at 10:45

Use only nextLine() method and the change the below lines of your code too for efficient

  Student1 s = new Student1();
    s.get(e ,f);

create constructor for Student1 class

  Student1(int c, String d)
    {
      a = c;
      b = d;
    }

Inside main() method

Student1 s = new Student1(e,f);

You can initialize the variables at the time of creating object by using constructor.

In Java Program Scanner is Working only once whereas i/ve used it twice

3 Answers3