Java

Character array convertion

Please support our Java advertiser: Programming Forums - DaniWeb Sister Site

Thread Solved

•

Join Date: Jan 2009

Posts: 20

Reputation: Teethous is an unknown quantity at this point

Solved Threads: 0

Teethous

Offline

Newbie Poster

Character array convertion

Oct 5th, 2009

[TEX]Hey everyone, I am having a problem with converting my character array. The program runs, but in the display box I get little squares instead of a string. The program should display the first char of the fist name, the first five of the second name and a random number between 10-99. Thanks much for your help.

 Help with Code Tags 
 Java Syntax (Toggle Plain Text) 
import java.util.Random;
import java.util.Scanner;
import javax.swing.JOptionPane;
 
 
public class Check {
 
	/**
	 * @param args
	 */
	public static void main(String[] args) 
	{
		Scanner input = new Scanner ( System.in); // creates an object of scanner
		Random random = new Random (); // creates an object of random
 
 
 
		char secondLength[]= new char [5]; // array for five characters
		int number = 0; // variable for number generated
		String s1 = new String( secondLength );
 
		//random number generator ranging from 10-99
		number = 10 + random.nextInt ( 90 );
 
		// gather data from user
		String firstName = JOptionPane.showInputDialog( "Enter first name" );
		String lastName = JOptionPane.showInputDialog( "Enter last name" );
 
		lastName.getChars ( 0, 5, secondLength, 0 );
 
		System.out.print( number );
		JOptionPane.showMessageDialog
		         ( null, "Your new string is: " + firstName.charAt(0) + s1 + number );	
 
 
	}
 
}
import java.util.Random;
import java.util.Scanner;
import javax.swing.JOptionPane;


public class Check {

	/**
	 * @param args
	 */
	public static void main(String[] args) 
	{
		Scanner input = new Scanner ( System.in); // creates an object of scanner
		Random random = new Random (); // creates an object of random
		
		

		char secondLength[]= new char [5]; // array for five characters
		int number = 0; // variable for number generated
		String s1 = new String( secondLength );
		
		//random number generator ranging from 10-99
		number = 10 + random.nextInt ( 90 );
		
		// gather data from user
		String firstName = JOptionPane.showInputDialog( "Enter first name" );
		String lastName = JOptionPane.showInputDialog( "Enter last name" );
		
		lastName.getChars ( 0, 5, secondLength, 0 );
		
		System.out.print( number );
		JOptionPane.showMessageDialog
		         ( null, "Your new string is: " + firstName.charAt(0) + s1 + number );	
			                                        			
		
	}

}

•

Join Date: Dec 2007

Posts: 1,711

Reputation: javaAddict is a name known to all

Solved Threads: 229

javaAddict javaAddict is online now

Online

Posting Virtuoso

Oct 6th, 2009

I believe that your problem is this:

 Help with Code Tags 
 Java Syntax (Toggle Plain Text) 
String s1 = new String( secondLength );
....
lastName.getChars ( 0, 5, secondLength, 0 );
String s1 = new String( secondLength );
....
lastName.getChars ( 0, 5, secondLength, 0 );

When you create the "s1" the "secondLength" array is empty. Then you call the method that gives to the array values, but the s1 was already created.
Try first to enter values to the array and then create the String "s1".

Also try to print the result at a System.out.println :

 Help with Code Tags 
 Java Syntax (Toggle Plain Text) 
String finalResult = firstName.charAt(0) + s1 + number;
System.out.println("Final Result :"+finalResult);
String finalResult = firstName.charAt(0) + s1 + number;
System.out.println("Final Result :"+finalResult);