How can I get the user input in Java?

I attempted to create a calculator, but I can not get it to work because I don't know how to get user input.

How can I get the user input in Java?

24 Answers
24

One of the simplest ways is to use a Scanner object as follows:

Scanner

import java.util.Scanner;

Scanner reader = new Scanner(System.in); // Reading from System.in
System.out.println("Enter a number: ");
int n = reader.nextInt(); // Scans the next token of the input as an int.
//once finished
reader.close();


Scanner reader1 = new Scanner(System.in);

You can use any of the following options based on the requirements.

Scanner

import java.util.Scanner;
Scanner scan = new Scanner(System.in);
String s = scan.next();
int i = scan.nextInt();


BufferedReader

InputStreamReader

import java.io.BufferedReader;
import java.io.InputStreamReader;
BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
String s = br.readLine();
int i = Integer.parseInt(br.readLine());


DataInputStream

import java.io.DataInputStream;
DataInputStream dis = new DataInputStream(System.in);
int i = dis.readInt();


The readLine method from the DataInputStream class has been deprecated. To get String value, you should use the previous solution with BufferedReader

readLine

DataInputStream

Console

import java.io.Console;
Console console = System.console();
String s = console.readLine();
int i = Integer.parseInt(console.readLine());


Apparently, this method does not work well in some IDEs.

DataInputStream

readInt

System.in

DataInput#readInt

DataInputStream

DataInput

You can use the Scanner class or the Console class

Console console = System.console();
String input = console.readLine("Enter input:");


You can get user input using BufferedReader.

BufferedReader

BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
String accStr;

System.out.println("Enter your Account number: ");
accStr = br.readLine();


It will store a String value in accStr so you have to parse it to an int using Integer.parseInt.

String

accStr

int

Integer.parseInt

int accInt = Integer.parseInt(accStr);


Here is how you can get the keyboard inputs:

Scanner scanner = new Scanner (System.in);
System.out.print("Enter your name");
name = scanner.next(); // Get what the user types.


You can make a simple program to ask for user's name and print what ever the reply use inputs.

Or ask user to enter two numbers and you can add, multiply, subtract, or divide those numbers and print the answers for user inputs just like a behavior of a calculator.

So there you need Scanner class. You have to import java.util.Scanner; and in the code you need to use

import java.util.Scanner;

Scanner input = new Scanner(System.in);


Input is a variable name.

Scanner input = new Scanner(System.in);

System.out.println("Please enter your name : ");
s = input.next(); // getting a String value

System.out.println("Please enter your age : ");
i = input.nextInt(); // getting an integer

System.out.println("Please enter your salary : ");
d = input.nextDouble(); // getting a double


See how this differs: input.next();, i = input.nextInt();, d = input.nextDouble();

input.next();

i = input.nextInt();

d = input.nextDouble();

According to a String, int and a double varies same way for the rest. Don't forget the import statement at the top of your code.

Also see the blog post "Scanner class and getting User Inputs".

To read a line or a string, you can use a BufferedReader object combined with an InputStreamReader one as follows:

BufferedReader

InputStreamReader

BufferedReader bufferReader = new BufferedReader(new InputStreamReader(System.in));
String inputLine = bufferReader.readLine();


Here, the program asks the user to enter a number. After that, the program prints the digits of the number and the sum of the digits.

import java.util.Scanner;

public class PrintNumber
public static void main(String args)
Scanner scan = new Scanner(System.in);
int num = 0;
int sum = 0;

System.out.println(
"Please enter a number to show its digits");
num = scan.nextInt();

System.out.println(
"Here are the digits and the sum of the digits");
while (num > 0)
System.out.println("==>" + num % 10);
sum += num % 10;
num = num / 10;

System.out.println("Sum is " + sum);




Use the System class to get the input.

System

http://fresh2refresh.com/java-tutorial/java-input-output/ :

We need three objects,

BufferedReader

InputStreamReader inp = new InputStreamReader(system.in);
BufferedReader br = new BufferedReader(inp);


System.in

S

Here is your program from the question using java.util.Scanner:

java.util.Scanner

import java.util.Scanner;

public class Example
public static void main(String args) 3. Divide

static int add(int lhs, int rhs) return lhs + rhs;
static int subtract(int lhs, int rhs) return lhs - rhs;
static int divide(int lhs, int rhs) return lhs / rhs;
static int multiply(int lhs, int rhs) return lhs * rhs;



Scanner

Just one extra detail. If you don't want to risk a memory/resource leak, you should close the scanner stream when you are finished:

myScanner.close();


Note that java 1.7 and later catch this as a compile warning (don't ask how I know that :-)

import java.util.Scanner;

class Daytwo
public static void main(String args)
System.out.println("HelloWorld");

Scanner reader = new Scanner(System.in);
System.out.println("Enter the number ");

int n = reader.nextInt();
System.out.println("You entered " + n);





Scanner input = new Scanner(System.in);
String inputval = input.next();


Add throws IOException beside main(), then

throws IOException

main()

DataInputStream input = new DataInputStream(System.in);
System.out.print("Enter your name");
String name = input.readLine();


It is very simple to get input in java, all you have to do is:

import java.util.Scanner;

class GetInputFromUser

public static void main(String args)

int a;
float b;
String s;

Scanner in = new Scanner(System.in);

System.out.println("Enter a string");
s = in.nextLine();
System.out.println("You entered string " + s);

System.out.println("Enter an integer");
a = in.nextInt();
System.out.println("You entered integer " + a);

System.out.println("Enter a float");
b = in.nextFloat();
System.out.println("You entered float " + b);




You can get user input like this using a BufferedReader:

InputStreamReader inp = new InputStreamReader(System.in);
BufferedReader br = new BufferedReader(inp);
// you will need to import these things.


This is how you apply them

String name = br.readline();


So when the user types in his name into the console, "String name" will store that information.

If it is a number you want to store, the code will look like this:

int x = Integer.parseInt(br.readLine());


Hop this helps!

import java.util.Scanner;

public class Myapplication
public static void main(String args)
Scanner in = new Scanner(System.in);
int a;
System.out.println("enter:");
a = in.nextInt();
System.out.println("Number is= " + a);




Can be something like this...

public static void main(String args)
Scanner reader = new Scanner(System.in);

System.out.println("Enter a number: ");
int i = reader.nextInt();
for (int j = 0; j < i; j++)
System.out.println("I love java");



This is a simple code that uses the System.in.read() function. This code just writes out whatever was typed. You can get rid of the while loop if you just want to take input once, and you could store answers in a character array if you so choose.

System.in.read()

package main;

import java.io.IOException;

public class Root

public static void main(String args)

new Root();


public Root()

while(true)

try

for(int y = 0; y < System.in.available(); ++y)

System.out.print((char)System.in.read());


catch(IOException ex)

ex.printStackTrace(System.out);
break;






I like the following:

public String readLine(String tPromptString)
byte tBuffer = new byte[256];
int tPos = 0;
System.out.print(tPromptString);

while(true)
byte tNextByte = readByte();
if(tNextByte == 10)
return new String(tBuffer, 0, tPos);


if(tNextByte != 13)
tBuffer[tPos] = tNextByte;
++tPos;





and for example, I would do:

String name = this.readLine("What is your name?")


Here is a more developed version of the accepted answer that addresses two common needs:

Code

package inputTest;

import java.util.Scanner;
import java.util.InputMismatchException;

public class InputTest
public static void main(String args)
Scanner reader = new Scanner(System.in);
System.out.println("Please enter integers. Type 0 to exit.");

boolean done = false;
while (!done)
System.out.print("Enter an integer: ");
try
int n = reader.nextInt();
if (n == 0)
done = true;

else
// do something with the input
System.out.println("tThe number entered was: " + n);


catch (InputMismatchException e)
System.out.println("tInvalid input type (must be an integer)");
reader.nextLine(); // Clear invalid input from scanner buffer.


System.out.println("Exiting...");
reader.close();




Example

Please enter integers. Type 0 to exit.
Enter an integer: 12
The number entered was: 12
Enter an integer: -56
The number entered was: -56
Enter an integer: 4.2
Invalid input type (must be an integer)
Enter an integer: but i hate integers
Invalid input type (must be an integer)
Enter an integer: 3
The number entered was: 3
Enter an integer: 0
Exiting...


Note that without nextLine(), the bad input will trigger the same exception repeatedly in an infinite loop. You might want to use next() instead depending on the circumstance, but know that input like this has spaces will generate multiple exceptions.

nextLine()

next()

this has spaces

import java.util.Scanner;

public class userinput
public static void main(String args)
Scanner input = new Scanner(System.in);

System.out.print("Name : ");
String name = input.next();
System.out.print("Last Name : ");
String lname = input.next();
System.out.print("Age : ");
byte age = input.nextByte();

System.out.println(" " );
System.out.println(" " );

System.out.println("Firt Name: " + name);
System.out.println("Last Name: " + lname);
System.out.println(" Age: " + age);




class ex1
public static void main(String args)
int a, b, c;
a = Integer.parseInt(args[0]);
b = Integer.parseInt(args[1]);
c = a + b;
System.out.println("c = " + c);


// Output
javac ex1.java
java ex1 10 20
c = 30


args

java ex1

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