How can a char be entered in Java from keyboard?
This question is related to
java
i found this way worked nice:
{
char [] a;
String temp;
Scanner keyboard = new Scanner(System.in);
System.out.println("please give the first integer :");
temp=keyboard.next();
a=temp.toCharArray();
}
you can also get individual one with String.charAt()
Using nextline and System.in.read as often proposed requires the user to hit enter after typing a character. However, people searching for an answer to this question, may also be interested in directly respond to a key press in a console!
I found a solution to do so using jline3, wherein we first change the terminal into rawmode to directly respond to keys, and then wait for the next entered character:
var terminal = TerminalBuilder.terminal()
terminal.enterRawMode()
var reader = terminal.reader()
var c = reader.read()
<dependency>
<groupId>org.jline</groupId>
<artifactId>jline</artifactId>
<version>3.12.3</version>
</dependency>
Here is a class 'getJ' with a static function 'chr()'. This function reads one char.
import java.io.InputStreamReader;
import java.io.BufferedReader;
import java.io.IOException;
class getJ {
static char chr()throws IOException{
BufferedReader bufferReader =new BufferedReader(new InputStreamReader(System.in));
return bufferReader.readLine().charAt(0);
}
}
In order to read a char use this:
anyFunc()throws IOException{
...
...
char c=getJ.chr();
}
Because of 'chr()' is static, you don't have to create 'getJ' by 'new' ; I mean you don't need to do:
getJ ob = new getJ;
c=ob.chr();
You should remember to add 'throws IOException' to the function's head. If it's impossible, use try / catch as follows:
anyFunc(){// if it's impossible to add 'throws IOException' here
...
try
{
char c=getJ.chr(); //reads a char into c
}
catch(IOException e)
{
System.out.println("IOException has been caught");
}
Credit to: tutorialspoint.com
See also: geeksforgeeks.
You can use a Scanner
for this. It's not clear what your exact requirements are, but here's an example that should be illustrative:
Scanner sc = new Scanner(System.in).useDelimiter("\\s*");
while (!sc.hasNext("z")) {
char ch = sc.next().charAt(0);
System.out.print("[" + ch + "] ");
}
If you give this input:
123 a b c x y z
The output is:
[1] [2] [3] [a] [b] [c] [x] [y]
So what happens here is that the Scanner
uses \s*
as delimiter, which is the regex for "zero or more whitespace characters". This skips spaces etc in the input, so you only get non-whitespace characters, one at a time.
Maybe you could try this code:
import java.io.*;
public class Test
{
public static void main(String[] args)
{
try
{
BufferedReader in = new BufferedReader(new InputStreamReader(System.in));
String userInput = in.readLine();
System.out.println("\n\nUser entered -> " + userInput);
}
catch(IOException e)
{
System.out.println("IOException has been caught");
}
}
}
You can use Scanner like so:
Scanner s= new Scanner(System.in);
char x = s.next().charAt(0);
By using the charAt function you are able to get the value of the first char without using external casting.
.... char ch; ... ch=scan.next().charAt(0); . . It's the easy way to get character.
Source: Stackoverflow.com