CODEWITHSUNDEEP
java
Gibbs
Gibbs

Reputation: 22956

How to read inputs from console

i have to read from the following format.

1

12

23

34

So, All inputs are separated by a line.

I have tried the following

   br = new Scanner(System.in);
   num = Integer.parseInt(br.next());
   while(br.hasNextInt()) {
        num = br.nextInt() ; 
        System.out.println(num);
    }

But it is not working as i expected. If i enter first input, it started processing it and prints. it is not waiting for me to enter next line. In C, i can make use of sscanf. but in java i have no idea how to allow user to enter multiline input? plese suggest some ideas

Upvotes: 3

Views: 358

Answers (8)

Biju CD
Biju CD

Reputation: 5109

 br = new BufferedReader(new InputStreamReader(System.in));
        int ch;
        while ((ch = br.read()) != -1) {
            if (ch == 10) {
                System.out.println();
            }
        }

Upvotes: 0

Mostafizar
Mostafizar

Reputation: 71

Scanner scanner = new.scanner(System.in);
String abc = scanner.next();
int a = scanner.nextInt();
scanner.close();

Upvotes: 0

Woodrow
Woodrow

Reputation: 136

Your question isn't entirely clear; however if you're trying to get the user input- then print it later, you can save the inputs to an ArrayList.

import java.util.ArrayList;
import java.util.Scanner;
public class tu {
    public static void main(String[] args) {
        Scanner in = new Scanner(System.in);
        System.out.println("Enter numbers, when you're done, enter a letter.");
        char firstLetter = '*';
        ArrayList<String> al = new ArrayList<>();
        while(!Character.isLetter(firstLetter)) {
            String input = "";
            boolean nullInput = false;
            do{
                if(nullInput)
                    System.out.println("Error; you can't enter an empty string.");
                nullInput = false;
                input = in.nextLine();
                if(input.length() == 0)
                    nullInput = true;
            }while(nullInput);
            firstLetter = input.charAt(0);
            al.add(input);
        }
        al.remove(al.size()-1);
        for(int i = 0; i < al.size(); i++) {
            System.out.println(); //creates newline in console
            System.out.println(al.get(i));

        }

    }


}

Upvotes: 0

Bruce
Bruce

Reputation: 8849

if you willing to read line by line . This code read line by line and exits when you input a empty line

br = new Scanner(System.in);
        int num=0;
        String input;
        input = br.nextLine();
        while(input.length()!=0) {
            num = Integer.parseInt(input);
            System.out.println(num);
            input = br.nextLine();
        }

Upvotes: 0

user2575725
user2575725

Reputation:

You must check for next available input and then get the input

br = new Scanner(System.in);
//num = Integer.parseInt(br.next());//remove this line
while(br.hasNextInt()) {//if number is avaialable
     num = br.nextInt(); //get that number
     System.out.println(num);
}

Below is sample code:

import java.util.Scanner;

class Ideone
{
    public static void main (String[] args)
    {
        Scanner sc = new Scanner(System.in);
        while(sc.hasNextInt())
            System.out.printf("input was: %d\n",sc.nextInt());
        sc.close();
    }
}

Upvotes: 1

TheRealFawby
TheRealFawby

Reputation: 112

It's not entirely obvious what you require , but this code ( below ) will allow you to read and store multiple lines of input , and then print them all when you're finished . 

systemInScanner = new Scanner ( System.in ) ;

ArrayList < String > arrayListOfStrings = new ArrayList < > ( ) ; 

while ( systemInScanner . hasNextLine ( ) ) 
{
    arrayListOfStrings . add ( systemInScanner . NextLine ( ) ) ; 
}

for ( String line : input ) 
{
    System . out . println ( line ) ; 
}

Upvotes: 0

argarevarg
argarevarg

Reputation: 31

Try

br = new Scanner(System.in); while (true) { int num = Integer.parseInt(br.nextLine()); System.out.println(num); }

Upvotes: 1

Siva Kumar
Siva Kumar

Reputation: 2006

   br = new Scanner(System.in);
   num = Integer.parseInt(br.next());
   br.nextLine();
     while(br.hasNextInt()) {
            num = br.nextInt() ; 
            br.nextLine();
            System.out.println(num);
        }

you need to add br.nextLine(); to start to read next line.

Upvotes: 0

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