Take a char input from the Scanner
JavaInputCharjava.util.scannerJava Problem Overview
I am trying to find a way to take a char
input from the keyboard.
I tried using:
Scanner reader = new Scanner(System.in);
char c = reader.nextChar();
This method doesn't exist.
I tried taking c
as a String
. Yet, it would not always work in every case, since the other method I am calling from my method requires a char
as an input. Therefore I have to find a way to explicitly take a char as an input.
Any help?
Java Solutions
Solution 1 - Java
You could take the first character from Scanner.next
:
char c = reader.next().charAt(0);
To consume exactly one character you could use:
char c = reader.findInLine(".").charAt(0);
To consume strictly one character you could use:
char c = reader.next(".").charAt(0);
Solution 2 - Java
Setup scanner:
reader.useDelimiter("");
After this reader.next()
will return a single-character string.
Solution 3 - Java
There is no API method to get a character from the Scanner. You should get the String using scanner.next()
and invoke String.charAt(0)
method on the returned String.
Scanner reader = new Scanner(System.in);
char c = reader.next().charAt(0);
Just to be safe with whitespaces you could also first call trim()
on the string to remove any whitespaces.
Scanner reader = new Scanner(System.in);
char c = reader.next().trim().charAt(0);
Solution 4 - Java
There are three ways to approach this problem:
-
Call
next()
on the Scanner, and extract the first character of the String (e.g.charAt(0)
) If you want to read the rest of the line as characters, iterate over the remaining characters in the String. Other answers have this code. -
Use
setDelimiter("")
to set the delimiter to an empty string. This will causenext()
to tokenize into strings that are exactly one character long. So then you can repeatedly callnext().charAt(0)
to iterate the characters. You can then set the delimiter to its original value and resume scanning in the normal way! -
Use the Reader API instead of the Scanner API. The
Reader.read()
method delivers a single character read from the input stream. For example:Reader reader = new InputStreamReader(System.in); int ch = reader.read(); if (ch != -1) { // check for EOF // we have a character ... }
When you read from the console via System.in
, the input is typically buffered by the operating system, and only "released" to the application when the user types ENTER. So if you intend your application to respond to individual keyboard strokes, this is not going to work. You would need to do some OS-specific native code stuff to turn off or work around line-buffering for console at the OS level.
Reference:
Solution 5 - Java
You can solve this problem, of "grabbing keyboard input one char at a time" very simply. Without having to use a Scanner all and also not clearing the input buffer as a side effect, by using this.
char c = (char)System.in.read();
If all you need is the same functionality as the C language "getChar()" function then this will work great. The Big advantage of the "System.in.read()" is the buffer is not cleared out after each char your grab. So if you still need all the users input you can still get the rest of it from the input buffer. The "char c = scanner.next().charAt(0);"
way does grab the char but will clear the buffer.
// Java program to read character without using Scanner
public class Main
{
public static void main(String[] args)
{
try {
String input = "";
// Grab the First char, also wait for user input if the buffer is empty.
// Think of it as working just like getChar() does in C.
char c = (char)System.in.read();
while(c != '\n') {
//<do your magic you need to do with the char here>
input += c; // <my simple magic>
//then grab the next char
c = (char)System.in.read();
}
//print back out all the users input
System.out.println(input);
} catch (Exception e){
System.out.println(e);
}
}
}
Hope this helpful, and good luck! P.S. Sorry i know this is an older post, but i hope that my answer bring new insight and could might help other people who also have this problem.
Solution 6 - Java
This actually doesn't work:
char c = reader.next().charAt(0);
There are some good explanations and references in this question: https://stackoverflow.com/questions/18746185/why-doesnt-the-scanner-class-have-a-nextchar-method "A Scanner breaks its input into tokens using a delimiter pattern", which is pretty open ended. For example when using this
c = lineScanner.next().charAt(0);
for this line of input "(1 + 9) / (3 - 1) + 6 - 2" the call to next returns "(1", c will be set to '(', and you'll end up losing the '1' on the next call to next()
Typically when you want to get a character you would like to ignore whitespace. This worked for me:
c = lineScanner.findInLine("[^\\s]").charAt(0);
Solution 7 - Java
The best way to take input of a character in Scanner class is:
Scanner sca=new Scanner(System.in);
System.out.println("enter a character");
char ch=sca.next().charAt(0);
Solution 8 - Java
You should use your custom input reader for faster results instead of extracting first character from reading String. Link for Custom ScanReader and explanation: https://gist.github.com/nik1010/5a90fa43399c539bb817069a14c3c5a8
Code for scanning Char :
BufferedInputStream br=new BufferedInputStream(System.in);
char a= (char)br.read();
Solution 9 - Java
import java.util.Scanner;
public class Test {
public static void main(String[] args) {
Scanner reader = new Scanner(System.in);
char c = reader.next(".").charAt(0);
}
}
To get only one character char c = reader.next(".").charAt(0);
Solution 10 - Java
There are two approaches, you can either take exactly one character or strictly one character. When you use exactly, the reader will take only the first character, irrespective of how many characters you input.
For example:
import java.util.Scanner;
public class ReaderExample {
public static void main(String[] args) {
try {
Scanner reader = new Scanner(System.in);
char c = reader.findInLine(".").charAt(0);
reader.close();
System.out.print(c);
} catch (Exception ex) {
System.out.println(ex.getMessage());
}
}
}
When you give a set of characters as input, say "abcd", the reader will consider only the first character i.e., the letter 'a'
But when you use strictly, the input should be just one character. If the input is more than one character, then the reader will not take the input
import java.util.Scanner;
public class ReaderExample {
public static void main(String[] args) {
try {
Scanner reader = new Scanner(System.in);
char c = reader.next(".").charAt(0);
reader.close();
System.out.print(c);
} catch (Exception ex) {
System.out.println(ex.getMessage());
}
}
}
Suppose you give input "abcd", no input is taken, and the variable c will have Null value.
Solution 11 - Java
try followings.
Scanner reader = new Scanner(System.in);
char c = reader.next().charAt(0);
this will get a character from the keyboard.
Solution 12 - Java
import java.util.Scanner;
public class userInput{
public static void main(String[] args){
// Creating your scanner with name kb as for keyBoard
Scanner kb = new Scanner(System.in);
String name;
int age;
char bloodGroup;
float height;
// Accepting Inputs from user
System.out.println("Enter Your Name");
name = kb.nextLine(); // for entire line of String including spaces
System.out.println("Enter Your Age");
age = kb.nextInt(); // for taking Int
System.out.println("Enter Your BloodGroup : A/B/O only");
bloodGroup = kb.next().charAt(0); // For character at position 0
System.out.println("Enter Your Height in Meters");
height = kb.nextFloat(); // for taking Float value
// closing your scanner object
kb.close();
// Outputting All
System.out.println("Name : " +name);
System.out.println("Age : " +age);
System.out.println("BloodGroup : " +bloodGroup);
System.out.println("Height : " +height+" m");
}
}
Solution 13 - Java
Try this: char c=S.nextLine().charAt(0);
Solution 14 - Java
You should get the String using scanner.next() and invoke String.charAt(0) method on the returned String.
Exmple :
import java.util.Scanner;
public class InputC{
public static void main(String[] args) {
// TODO Auto-generated method stub
// Declare the object and initialize with
// predefined standard input object
Scanner scanner = new Scanner(System.in);
System.out.println("Enter a character: ");
// Character input
char c = scanner.next().charAt(0);
// Print the read value
System.out.println("You have entered: "+c);
}
}
output
Enter a character:
a
You have entered: a
Solution 15 - Java
you just need to write this for getting value in char type.
char c = reader.next().charAt(0);
Solution 16 - Java
Simple solution to read a charachter from user input. Read a String. Then use charAt(0) over String
Scanner reader = new Scanner(System.in);
String str = reader.next();
char c = str.charAt(0);
That's it.
Solution 17 - Java
Just use...
Scanner keyboard = new Scanner(System.in);
char c = keyboard.next().charAt(0);
This gets the first character of the next input.
Solution 18 - Java
// Use a BufferedReader to read characters from the console.
import java.io.*;
class BRRead {
public static void main(String args[]) throws IOException
{
char c;
BufferedReader br = new
BufferedReader(new InputStreamReader(System.in));
System.out.println("Enter characters, 'q' to quit.");
// read characters
do {
c = (char) br.read();
System.out.println(c);
} while(c != 'q');
}
}
Solution 19 - Java
import java.io.*;
class abc // enter class name (here abc is class name)
{
public static void main(String arg[])
throws IOException // can also use Exception
{
BufferedReader z =
new BufferedReader(new InputStreamReader(System.in));
char ch = (char) z.read();
} // PSVM
} // class
Solution 20 - Java
Try this
Scanner scanner=new Scanner(System.in);
String s=scanner.next();
char c=s.charAt(0);
Solution 21 - Java
Scanner key = new Scanner(System.in);
//shortcut way
char firstChar=key.next().charAt(0);
//how it works;
/*key.next() takes a String as input then,
charAt method is applied on that input (String)
with a parameter of type int (position) that you give to get
that char at that position.
You can simply read it out as:
the char at position/index 0 from the input String
(through the Scanner object key) is stored in var. firstChar (type char) */
//you can also do it in a bit elabortive manner to understand how it exactly works
String input=key.next(); // you can also write key.nextLine to take a String with spaces also
char firstChar=input.charAt(0);
char charAtAnyPos= input.charAt(pos); // in pos you enter that index from where you want to get the char from
By the way, you can't take a char directly as an input. As you can see above, a String is first taken then the charAt(0); is found and stored
Solution 22 - Java
You could use typecasting:
Scanner sc= new Scanner(System.in);
char a=(char) sc.next();
This way you will take input in String due to the function 'next()' but then it will be converted into character due to the 'char' mentioned in the brackets.
This method of conversion of data type by mentioning the destination data type in brackets is called typecating. It works for me, I hope it works for u :)
Solution 23 - Java
To find the index of a character in a given sting, you can use this code:
package stringmethodindexof;
import java.util.Scanner;
import javax.swing.JOptionPane;
/**
*
* @author ASUS//VERY VERY IMPORTANT
*/
public class StringMethodIndexOf {
/**
* @param args the command line arguments
*/
public static void main(String[] args) {
// TODO code application logic here
String email;
String any;
//char any;
//any=JOptionPane.showInputDialog(null,"Enter any character or string to find out its INDEX NUMBER").charAt(0);
//THE AVOBE LINE IS FOR CHARACTER INPUT LOL
//System.out.println("Enter any character or string to find out its INDEX NUMBER");
//Scanner r=new Scanner(System.in);
// any=r.nextChar();
email = JOptionPane.showInputDialog(null,"Enter any string or anything you want:");
any=JOptionPane.showInputDialog(null,"Enter any character or string to find out its INDEX NUMBER");
int result;
result=email.indexOf(any);
JOptionPane.showMessageDialog(null, result);
}
}
Solution 24 - Java
The easiest way is, first change the variable to a String and accept the input as a string. Then you can control based on the input variable with an if-else or switch statement as follows.
Scanner reader = new Scanner(System.in);
String c = reader.nextLine();
switch (c) {
case "a":
<your code here>
break;
case "b":
<your code here>
break;
default:
<your code here>
}