Java: How to get input from System.console()
Asked Answered
M

9

176

I am trying to use Console class to get input from user but a null object is returned when I call System.console(). Do I have to change anything before using System.console?

Console co=System.console();
System.out.println(co);
try{
    String s=co.readLine();
}
Marsha answered 10/1, 2011 at 7:1 Comment(4)
is this for android ? (i'm guessing from your user id)Morphophonemics
Are you using eclipse to start your program? Try to start your program without eclipse using java.exe.Teach
Have a look at McDowell's project "AbstractingTheJavaConsole": illegalargumentexception.googlecode.com/svn/trunk/code/java/…Stratocumulus
@RyanFernandes How is his name relevant to his question?Unsecured
S
255

Using Console to read input (usable only outside of an IDE):

System.out.print("Enter something:");
String input = System.console().readLine();

Another way (works everywhere):

import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;

public class Test {
    public static void main(String[] args) throws IOException { 
        BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
        System.out.print("Enter String");
        String s = br.readLine();
        System.out.print("Enter Integer:");
        try {
            int i = Integer.parseInt(br.readLine());
        } catch(NumberFormatException nfe) {
            System.err.println("Invalid Format!");
        }
    }
}

System.console() returns null in an IDE.
So if you really need to use System.console(), read this solution from McDowell.

Stratocumulus answered 10/1, 2011 at 9:16 Comment(4)
why we need BufferedReader to read the input why cant we directly read from InputStreamReaderBimah
Got the answer: With a BufferedInputStream, the method delegates to an overloaded read() method that reads 8192 amount of bytes and buffers them until they are needed. It still returns only the single byte (but keeps the others in reserve). This way the BufferedInputStream makes less native calls to the OS to read from the file. ThanksBimah
In case we want to read password from user, #22546103 masks the line with asterisk.Herat
@Bimah another reason might be that BufferedReader provides the method readLine() which does not exist for InputStreamReader.Aggarwal
S
123
Scanner in = new Scanner(System.in);

int i = in.nextInt();
String s = in.next();
Speaker answered 10/1, 2011 at 7:11 Comment(5)
However nextLine() is very messy to work with. At best it will give you a headache when trying to get whole lines from the console.Duenas
@Yokhen could you please given an example where in.nextLine() would create problems?Gerundive
(1) By default the delimiter of Scanner is space, so when user enters multiple texts, it will cause the software to proceed to next few next() and the logic in our software will be wrong. (2) If I use nextLine() to read the whole full sentence which including space and \n\r, I need to trim() user input. (3) next() will wait for user input but nextLine() will not. (4) I tested useDelimiter("\\r\\n"), but it causes the next() logic in our software somewhere else to be wrong again. Conclusion, it is indeed quite messy to use Scanner to read user input. BufferedReader is the best.Herat
I also have problems with scanner, notably I get a java.util.NoSuchElementException that I don't quite understand.Tomasatomasina
All this should be inside a try-with-resources and the method should be in.nextLine() if you want to read a line.Damara
K
36

There are few ways to read input string from your console/keyboard. The following sample code shows how to read a string from the console/keyboard by using Java.

public class ConsoleReadingDemo {

public static void main(String[] args) {

    // ====
    BufferedReader reader = new BufferedReader(new InputStreamReader(System.in));
    System.out.print("Please enter user name : ");
    String username = null;
    try {
        username = reader.readLine();
    } catch (IOException e) {
        e.printStackTrace();
    }
    System.out.println("You entered : " + username);

    // ===== In Java 5, Java.util,Scanner is used for this purpose.
    Scanner in = new Scanner(System.in);
    System.out.print("Please enter user name : ");
    username = in.nextLine();      
    System.out.println("You entered : " + username);


    // ====== Java 6
    Console console = System.console();
    username = console.readLine("Please enter user name : ");   
    System.out.println("You entered : " + username);

}
}

The last part of code used java.io.Console class. you can not get Console instance from System.console() when running the demo code through Eclipse. Because eclipse runs your application as a background process and not as a top-level process with a system console.

Kirksey answered 28/8, 2014 at 6:12 Comment(0)
V
18

It will depend on your environment. If you're running a Swing UI via javaw for example, then there isn't a console to display. If you're running within an IDE, it will very much depend on the specific IDE's handling of console IO.

From the command line, it should be fine though. Sample:

import java.io.Console;

public class Test {

    public static void main(String[] args) throws Exception {
        Console console = System.console();
        if (console == null) {
            System.out.println("Unable to fetch console");
            return;
        }
        String line = console.readLine();
        console.printf("I saw this line: %s", line);
    }
}

Run this just with java:

> javac Test.java
> java Test
Foo  <---- entered by the user
I saw this line: Foo    <---- program output

Another option is to use System.in, which you may want to wrap in a BufferedReader to read lines, or use Scanner (again wrapping System.in).

Vola answered 10/1, 2011 at 7:8 Comment(0)
I
7

Found some good answer here regarding reading from console, here another way use 'Scanner' to read from console:

import java.util.Scanner;
String data;

Scanner scanInput = new Scanner(System.in);
data= scanInput.nextLine();

scanInput.close();            
System.out.println(data);
Intelligent answered 27/3, 2014 at 4:40 Comment(2)
You may not want to call close() the scanner in this case, because it will close System.in and prevent your application from reading from it later on. (Reading later throws error "no line found")Froze
Yes, I agree to that, close() should not be use if the intention is to use System.in later in the program.Intelligent
G
5

Try this. hope this will help.

    String cls0;
    String cls1;

    Scanner in = new Scanner(System.in);  
    System.out.println("Enter a string");  
    cls0 = in.nextLine();  

    System.out.println("Enter a string");  
    cls1 = in.nextLine(); 
Gris answered 25/9, 2016 at 16:36 Comment(0)
T
3

The following takes athspk's answer and makes it into one that loops continually until the user types "exit". I've also written a followup answer where I've taken this code and made it testable.

import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;

public class LoopingConsoleInputExample {

   public static final String EXIT_COMMAND = "exit";

   public static void main(final String[] args) throws IOException {
      BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
      System.out.println("Enter some text, or '" + EXIT_COMMAND + "' to quit");

      while (true) {

         System.out.print("> ");
         String input = br.readLine();
         System.out.println(input);

         if (input.length() == EXIT_COMMAND.length() && input.toLowerCase().equals(EXIT_COMMAND)) {
            System.out.println("Exiting.");
            return;
         }

         System.out.println("...response goes here...");
      }
   }
}

Example output:

Enter some text, or 'exit' to quit
> one
one
...response goes here...
> two
two
...response goes here...
> three
three
...response goes here...
> exit
exit
Exiting.
Telescopic answered 13/10, 2015 at 15:37 Comment(0)
S
3

I wrote the Text-IO library, which can deal with the problem of System.console() being null when running an application from within an IDE.

It introduces an abstraction layer similar to the one proposed by McDowell. If System.console() returns null, the library switches to a Swing-based console.

In addition, Text-IO has a series of useful features:

  • supports reading values with various data types.
  • allows masking the input when reading sensitive data.
  • allows selecting a value from a list.
  • allows specifying constraints on the input values (format patterns, value ranges, length constraints etc.).

Usage example:

TextIO textIO = TextIoFactory.getTextIO();

String user = textIO.newStringInputReader()
        .withDefaultValue("admin")
        .read("Username");

String password = textIO.newStringInputReader()
        .withMinLength(6)
        .withInputMasking(true)
        .read("Password");

int age = textIO.newIntInputReader()
        .withMinVal(13)
        .read("Age");

Month month = textIO.newEnumInputReader(Month.class)
        .read("What month were you born in?");

textIO.getTextTerminal().println("User " + user + " is " + age + " years old, " +
        "was born in " + month + " and has the password " + password + ".");

In this image you can see the above code running in a Swing-based console.

Strengthen answered 14/11, 2016 at 7:47 Comment(0)

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