I have the user entering a single character into the program and it is stored as a string. I would like to know how I could check to see if the character that was entered is a letter or a digit. I have an if statement, so if its a letter its prints that it's a letter, and the same for a digit. The code I have so far doesn't work but I feel like I'm close. Any help you can offer is appreciated.
System.out.println("Please enter a single character: ");
String character = in.next();
System.out.println(character);
if (character.isLetter()){
System.out.println("The character entered is a letter.");
}
else (character.isDigit()){
Syste.out.println("The character entered is a digit.");
This question is related to
java
I have coded a sample program that checks if a string contains a number in it! I guess it will serve for this purpose as well.
public class test {
public static void main(String[] args) {
String c;
boolean b;
System.out.println("Enter the value");
Scanner s = new Scanner(System.in);
c = s.next();
b = containsNumber(c);
try {
if (b == true) {
throw new CharacterFormatException();
} else {
System.out.println("Valid String \t" + c);
}
} catch (CharacterFormatException ex) {
System.out.println("Exception Raised-Contains Number");
}
}
static boolean containsNumber(String c) {
char[] ch = new char[10];
ch = c.toCharArray();
for (int i = 0; i < ch.length; i++) {
if ((ch[i] >= 48) && (ch[i] <= 57)) {
return true;
}
}
return false;
}
}
CharacterFormatException
is a user defined Exception. Suggest me if any changes can be made.
char charInt=character.charAt(0);
if(charInt>=48 && charInt<=57){
System.out.println("not character");
}
else
System.out.println("Character");
Look for ASCII table to see how the int value are hardcoded .
You need to convert your string into character..
String character = in.next();
char myChar = character.charAt(0);
if (Character.isDigit(myChar)) {
// print true
}
Check Character for other methods..
This is a little tricky, the value you enter at keyboard, is a String value, so you have to pitch the first character with method line.chartAt(0)
where, 0 is the index of the first character, and store this value in a char variable as in char c= line.charAt(0)
now with the use of method isDigit()
and isLetter()
from class Character
you can differentiate between a Digit and Letter.
here is a code for your program:
import java.util.Scanner;
class Practice
{
public static void main(String[] args)
{
Scanner in = new Scanner(System.in);
System.out.println("Input a letter");
String line = in.nextLine();
char c = line.charAt(0);
if( Character.isDigit(c))
System.out.println(c +" Is a digit");
else if (Character.isLetter(c))
System.out.println(c +" Is a Letter");
}
}
You could use the existing methods from the Character class. Take a look at the docs:
http://download.java.net/jdk7/archive/b123/docs/api/java/lang/Character.html#isDigit(char)
So, you could do something like this...
String character = in.next();
char c = character.charAt(0);
...
if (Character.isDigit(c)) {
...
} else if (Character.isLetter(c)) {
...
}
...
If you ever want to know exactly how this is implemented, you could always look at the Java source code.
char temp = yourString.charAt(0);
if(Character.isDigit(temp))
{
..........
}else if (Character.isLetter(temp))
{
......
}else
{
....
}
import java.util.*;
public class String_char
{
public static void main(String arg[]){
Scanner in = new Scanner(System.in);
System.out.println("Enter the value");
String data;
data = in.next();
int len = data.length();
for (int i = 0 ; i < len ; i++){
char ch = data.charAt(i);
if ((ch >= '0' && ch <= '9')){
System.out.println("Number ");
}
else if((ch >= 'a' && ch <= 'z') || (ch >= 'A' && ch <= 'Z')){
System.out.println("Character");
}
else{
System.out.println("Symbol");
}
}
}
}
This is the way how to check whether a given character is alphabet or not
public static void main(String[] args) {
Scanner sc = new Scanner(System.in);
char c = sc.next().charAt(0);
if((c >= 'a' && c <= 'z') || (c >= 'A' && c <= 'Z'))
System.out.println(c + " is an alphabet.");
else
System.out.println(c + " is not an alphabet.");
}
By using regular expressions:
boolean isChar = character.matches("[a-zA-z]{1}");
boolean isDigit = character.matches("\\d{1}");
Ummm, you guys are forgetting the Character.isLetterOrDigit
method:
boolean x;
String character = in.next();
char c = character.charAt(0);
if(Character.isLetterOrDigit(charAt(c)))
{
x = true;
}
You could use:
if (Character.isLetter(character.charAt(0))){
....
You could do this by Regular Expression as follows you could use this code
EditText et = (EditText) findViewById(R.id.editText);
String NumberPattern = "[0-9]+";
String Number = et.getText().toString();
if (Number.matches(NumberPattern) && s.length() > 0)
{
//code for number
}
else
{
//code for incorrect number pattern
}
Source: Stackoverflow.com