How to get the user input in Java?

Java Problem Overview

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?

Java Solutions

Solution 1 - Java

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

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();

Solution 2 - Java

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

`Scanner` class

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

`BufferedReader` and `InputStreamReader` classes

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

`DataInputStream` class

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

`Console` class

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.

Solution 3 - Java

You can use the Scanner class or the Console class

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

Solution 4 - Java

You can get user input using 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.

int accInt = Integer.parseInt(accStr);

Solution 5 - Java

Here is how you can get the keyboard inputs:

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

Solution 6 - Java

The best two options are BufferedReader and Scanner.

The most widely used method is Scanner and I personally prefer it because of its simplicity and easy implementation, as well as its powerful utility to parse text into primitive data.

Advantages of Using Scanner

Easy to use the Scanner class
Easy input of numbers (int, short, byte, float, long and double)
Exceptions are unchecked which is more convenient. It is up to the programmer to be civilized, and specify or catch the exceptions.
Is able to read lines, white spaces, and regex-delimited tokens

Advantages of BufferedInputStream

BufferedInputStream is about reading in blocks of data rather than a single byte at a time
Can read chars, char arrays, and lines
Throws checked exceptions
Fast performance
Synchronized (you cannot share Scanner between threads)

Overall each input method has different purposes.

If you are inputting large amount of data BufferedReader might be better for you
If you are inputting lots of numbers Scanner does automatic parsing which is very convenient

For more basic uses I would recommend the Scanner because it is easier to use and easier to write programs with. Here is a quick example of how to create a Scanner. I will provide a comprehensive example below of how to use the Scanner

Scanner scanner = new Scanner (System.in); // create scanner
System.out.print("Enter your name");       // prompt user
name = scanner.next();                     // get user input

(For more info about BufferedReader see How to use a BufferedReader and see Reading lines of Chars)

java.util.Scanner

import java.util.InputMismatchException; // import the exception catching class
import java.util.Scanner; // import the scanner class

public class RunScanner {

    // main method which will run your program
    public static void main(String args[]) {

        // create your new scanner
        // Note: since scanner is opened to "System.in" closing it will close "System.in". 
        // Do not close scanner until you no longer want to use it at all.
        Scanner scanner = new Scanner(System.in);

        // PROMPT THE USER
        // Note: when using scanner it is recommended to prompt the user with "System.out.print" or "System.out.println"
        System.out.println("Please enter a number");

        // use "try" to catch invalid inputs
        try {

            // get integer with "nextInt()"
            int n = scanner.nextInt();
                            
            
            System.out.println("Please enter a decimal"); // PROMPT
            // get decimal with "nextFloat()"
            float f = scanner.nextFloat();

            
            System.out.println("Please enter a word"); // PROMPT
            // get single word with "next()"
            String s = scanner.next();
            
            // ---- Note: Scanner.nextInt() does not consume a nextLine character /n 
            // ---- In order to read a new line we first need to clear the current nextLine by reading it:
            scanner.nextLine(); 
            // ----
            System.out.println("Please enter a line"); // PROMPT
            // get line with "nextLine()"
            String l = scanner.nextLine();


            // do something with the input
            System.out.println("The number entered was: " + n);
            System.out.println("The decimal entered was: " + f);
            System.out.println("The word entered was: " + s);
            System.out.println("The line entered was: " + l);
            
            
        }
        catch (InputMismatchException e) {
            System.out.println("\tInvalid input entered. Please enter the specified input");
        }

        scanner.close(); // close the scanner so it doesn't leak
    }
}

^{Note: Other classes such as Console and DataInputStream are also viable alternatives.}

^{Console has some powerful features such as ability to read passwords, however, is not available in all IDE's (such as Eclipse). The reason this occurs is because Eclipse runs your application as a background process and not as a top-level process with a system console. Here is a link to a useful example on how to implement the Console class.}

^{DataInputStream is primarily used for reading input as a primitive datatype, from an underlying input stream, in a machine-independent way. DataInputStream is usually used for reading binary data. It also provides convenience methods for reading certain data types. For example, it has a method to read a UTF String which can contain any number of lines within them.}

^{However, it is a more complicated class and harder to implement so not recommended for beginners. Here is a link to a useful example how to implement a DataInputStream.}

Solution 7 - Java

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

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();

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".

Solution 8 - Java

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

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

Solution 9 - Java

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);            
    }
}

Solution 10 - Java

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

import java.util.Scanner;

public class Example {
    public static void main(String[] args) {
        int input = 0;
        System.out.println("The super insano calculator");
        System.out.println("enter the corrosponding number:");
        Scanner reader3 = new Scanner(System.in);
        System.out.println(
            "1. Add | 2. Subtract | 3. Divide | 4. Multiply");

        input = reader3.nextInt();

        int a = 0, b = 0;

        Scanner reader = new Scanner(System.in);
        System.out.println("Enter the first number");
        // get user input for a
        a = reader.nextInt();

        Scanner reader1 = new Scanner(System.in);
        System.out.println("Enter the scend number");
        // get user input for b
        b = reader1.nextInt();

        switch (input){
            case 1:  System.out.println(a + " + " + b + " = " + add(a, b));
                     break;
            case 2:  System.out.println(a + " - " + b + " = " + subtract(a, b));
                     break;
            case 3:  System.out.println(a + " / " + b + " = " + divide(a, b));
                     break;
            case 4:  System.out.println(a + " * " + b + " = " + multiply(a, b));
                     break;
            default: System.out.println("your input is invalid!");
                     break;
        }
    }
    
    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; }
}

Solution 11 - Java

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

Solution 12 - Java

Scanner input=new Scanner(System.in);
int integer=input.nextInt();
String string=input.next();
long longInteger=input.nextLong();

Solution 13 - Java

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 :-)

Solution 14 - Java

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

Collecting user input repeatedly until an exit value has been entered
Dealing with invalid input values (non-integers in this example)

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.

Solution 15 - Java

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);
		
	}
}

Solution 16 - Java

Add throws IOException beside main(), then

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

Solution 17 - Java

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);
    }
}

Solution 18 - Java

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);
     }
}

Solution 19 - Java

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");
}

Solution 20 - Java

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!

Solution 21 - Java

You can get the user input using Scanner. You can use the proper input validation using proper methods for different data types like next() for String or nextInt() for Integer.

import java.util.Scanner;

Scanner scanner = new Scanner(System.in);

//reads the input until it reaches the space
System.out.println("Enter a string: ");
String str = scanner.next();
System.out.println("str = " + str);

//reads until the end of line
String aLine = scanner.nextLine();

//reads the integer
System.out.println("Enter an integer num: ");
int num = scanner.nextInt();
System.out.println("num = " + num);

//reads the double value
System.out.println("Enter a double: ");
double aDouble = scanner.nextDouble();
System.out.println("double = " + aDouble);


//reads the float value, long value, boolean value, byte and short
double aFloat = scanner.nextFloat();
long aLong = scanner.nextLong();
boolean aBoolean = scanner.nextBoolean();
byte aByte = scanner.nextByte();
short aShort = scanner.nextShort();

scanner.close();

Solution 22 - Java

import java.util.Scanner;

public class Main {
    public static void main(String[] args) {
        System.out.println("Welcome to the best program in the world! ");
        while (true) {
            System.out.print("Enter a query: ");
            Scanner scan = new Scanner(System.in);
            String s = scan.nextLine();
            if (s.equals("q")) {
                System.out.println("The program is ending now ....");
                break;
            } else  {
                System.out.println("The program is running...");
            }
        }
    }
}

Solution 23 - 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.

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;
			}
		}
	}	
}

Solution 24 - Java

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?")

Solution 25 - Java

Keyboard entry using Scanner is possible, as others have posted. But in these highly graphic times it is pointless making a calculator without a graphical user interface (GUI).

In modern Java this means using a JavaFX drag-and-drop tool like Scene Builder to lay out a GUI that resembles a calculator's console. Note that using Scene Builder is intuitively easy and demands no additional Java skill for its event handlers that what you already may have.

For user input, you should have a wide TextField at the top of the GUI console.

This is where the user enters the numbers that they want to perform functions on. Below the TextField, you would have an array of function buttons doing basic (i.e. add/subtract/multiply/divide and memory/recall/clear) functions. Once the GUI is lain out, you can then add the 'controller' references that link each button function to its Java implementation, e.g a call to method in your project's controller class.

This video is a bit old but still shows how easy Scene Builder is to use.

Solution 26 - Java

The most simple way to get user input would be to use Scanner. Here's an example of how it's supposed to be used:

import java.util.Scanner;
public class main {
public static void main(String[]args) {
Scanner sc=new Scanner(System.in);
int a;
String b;
System.out.println("Type an integer here: ");
a=sc.nextInt();
System.out.println("Type anything here:");
b=sc.nextLine();

The line of code import java.util.Scanner; tells the program that the programmer will be using user inputs in their code. Like it says, it imports the scanner utility. Scanner sc=new Scanner(System.in); tells the program to start the user inputs. After you do that, you must make a string or integer without a value, then put those in the line a=sc.nextInt(); or a=sc.nextLine();. This gives the variables the value of the user inputs. Then you can use it in your code. Hope this helps.

Solution 27 - Java

Using JOptionPane you can achieve it.

Int a =JOptionPane.showInputDialog(null,"Enter number:");

Solution 28 - Java

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);
    }
}

Solution 29 - Java

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

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How to get the user input in Java?

Java Problem Overview

Java Solutions

Solution 1 - Java

Solution 2 - Java

`Scanner` class

`BufferedReader` and `InputStreamReader` classes

`DataInputStream` class

`Console` class

Solution 3 - Java

Solution 4 - Java

Solution 5 - Java

Solution 6 - Java

java.util.Scanner

Solution 7 - Java

Solution 8 - Java

Solution 9 - Java

Solution 10 - Java

Solution 11 - Java

Solution 12 - Java

Solution 13 - Java

Solution 14 - Java

Solution 15 - Java

Solution 16 - Java

Solution 17 - Java

Solution 18 - Java

Solution 19 - Java

Solution 20 - Java

Solution 21 - Java

Solution 22 - Java

Solution 23 - Java

Solution 24 - Java

Solution 25 - Java

Solution 26 - Java

Solution 27 - Java

Solution 28 - Java

Solution 29 - Java

Order Bars in ggplot2 bar graph

How to capture a list of specific type with mockito

Attributions

Java Problem Overview

Java Solutions

Solution 1 - Java

Solution 2 - Java

Scanner class

BufferedReader and InputStreamReader classes

DataInputStream class

Console class

Solution 3 - Java

Solution 4 - Java

Solution 5 - Java

Solution 6 - Java

java.util.Scanner

Solution 7 - Java

Solution 8 - Java

Solution 9 - Java

Solution 10 - Java

Solution 11 - Java

Solution 12 - Java

Solution 13 - Java

Solution 14 - Java

Solution 15 - Java

Solution 16 - Java

Solution 17 - Java

Solution 18 - Java

Solution 19 - Java

Solution 20 - Java

Solution 21 - Java

Solution 22 - Java

Solution 23 - Java

Solution 24 - Java

Solution 25 - Java

Solution 26 - Java

Solution 27 - Java

Solution 28 - Java

Solution 29 - Java

Order Bars in ggplot2 bar graph

How to capture a list of specific type with mockito

Attributions

`Scanner` class

`BufferedReader` and `InputStreamReader` classes

`DataInputStream` class

`Console` class