[Riprickwright]

My Project for my java class is to create a phone book with certain keys having a certain function. My problem is with the entry function because my instructor wants us to put the inputs into an object array that can hold no more than 100 values. The last fix I made on this ran with gave me something along the lines of "Entry@14318bb
Entry@ca0b6
Entry@10b30a7" for a hundred lines instead of what I put in.

import java.io.*;
import java.util.*;

class Entry {
	public String name, number, note;	
	
	
	
	Entry()
	{
		name = "";
		number = "";
		note = "";
	}
}





public class Phonebook {
	public static Entry[] entrylist = new Entry[100];
	


	public static void main(String[] args) throws Exception {
		for (int i=0; i<entrylist.length; i++){
			entrylist[i] = new Entry();
		}
		Scanner input = new Scanner(System.in);
		File file = new File("contactlist.txt");
		BufferedWriter output = new BufferedWriter(new FileWriter(file));
		System.out.println("Codes are entered as 1 to 8 characters. \nUse \"e\" for enter, \"f\" for find, \"l\" to list, \"q\" to quit.");
		int i = 0;
		int j = 0;
		while (i == 0) {
			System.out.println("Command: ");
			String g = input.next();
			String[] token = g.split(" ");
			String k = token[0];
			if (k.equals("q")) {
				break;
			}
			if (k.equals("e")) {
				if (j >= entrylist.length){
					System.out.println("You have already used all of your entries...");
				}
				else{
				entrylist[j].name = input.next();
				System.out.println("Enter number: ");
				entrylist[j].number = input.next();
				System.out.println("Note: ");
				entrylist[j].note = input.next();
				j++;
				}
			}
			if (k.equals("f")) {
				System.out.println("Ha, im not going to find shit!");
			}
			if (k.equals("l")) {
				for (int l = 0; l < entrylist.length; l++){
					if (entrylist == null){
						System.out.println("");
					}
					else
					System.out.println(entrylist[l]);
				}
			}
		}
		}


}

[Riprickwright]

* I used the function "e" to enter the name number and note a few times then used "l" to display to get the hundred lines of "Entry@14318bb
Entry@ca0b6
Entry@10b30a7"

[pbl]

System.out.println(entrylist[l]);

your Entry class will need a toString() method something like

public String toString() {
   return name + " " + number + " " + note;
}

[oha055]

You should override the Entry class' toString() method.

public String toString() {
return name + " " + number + " " + note;
}

System.out.println(entrylist[l].toString());

[Silent Ace]

Also break is not terminating the console it only breaks the code.

if (k.equals("q")) {
     System.exit(0);
}

[Riprickwright]

Thank you guys so much. I have one more question, how would I go about alphabetizing entrylist by the first letter so that entrylist[0] would be the first name in alphabetical order?
I have a rough idea:

if (k.equals("l")) {
        for (int a = 0; a < entrylist.length; a++){
	String[] t = entrylist[a].name.split("");
	String kd = t[0];
	}

Would this be how I split the first letter from the list and how would I arrange them using it?

[pbl]

Simply implements Comparable in your Entry class

class Entry implements Comparable<Entry>{
	public String name, number, note;	
		
	Entry()
	{
		name = "";
		number = "";
		note = "";
	}

        public String toString() {
           return name + " " + number + " " + note;
        }

        public int compareTo(Entry other) {
           return name.compareTo(other.name);
        }
}

Then in your code you can simply: Arrays.sort(entryList);

Happy coding


*Edited: oups forgot to remove extra code

This post has been edited by pbl: 11 April 2012 - 07:25 PM

[Riprickwright]

Where should I put that at? I put it the if "l" statement:

if (k.equals("l")) {
				for (int a = 0; a < entrylist.length; a++){
					Arrays.sort(entrylist);						
					}
				for (int l = 0; l < entrylist.length; l++){
					if (entrylist == null){
						System.out.println("");
					}
					else
					System.out.println(entrylist[l].toString());
				}
			}

and when I entered "l" I got:
Exception in thread "main" java.lang.ClassCastException: Entry cannot be cast to java.lang.Comparable
at java.util.Arrays.mergeSort(Unknown Source)
at java.util.Arrays.mergeSort(Unknown Source)
at java.util.Arrays.mergeSort(Unknown Source)
at java.util.Arrays.mergeSort(Unknown Source)
at java.util.Arrays.mergeSort(Unknown Source)
at java.util.Arrays.sort(Unknown Source)
at Phonebook.main(Phonebook.java:69)
And line 69 is where I put the statement at

[pbl]

Arrays.sort(entrylist);

just once do need to put it in a loop

Do you still have null entries in your array[] or they are all filled ?
Did you add the implements Comparable<Entry> in your Entry class declaration ?

[Riprickwright]

Entry()
	{
		name = "";
		number = "";
		note = "" ;
	}
	public String toString() {		
		return name + " " + number + " " + note;
	}
	public int compareTo(Entry other) {
		return name.compareTo(other.name);
	}
	}

I did and I changed the "l" so there is no loop but I still get the same result.

[pbl]

Do you still have null entries in your array[] or they are all filled ?
Did you add the implements Comparable<Entry> in your Entry class declaration ?

class Entry implements Comparable<Entry> {

[Riprickwright]

Sorry about that. Yes all of the statements are no longer null and this works out now, but when I use the "l" statement I get all of the other blank entries:
"











name a
name b"
how would I get rid of them in the print statement?
Thank you so much for your patience with me so far.

[pbl]

Yes because you filled you array with new Entry() where name and num are ""

you can always

Arrays.sort(entries);
for(int i = 0; i < entries.length; i++) {
   String str = entries[i].trim();
   if(str.length() > 0)
      System.out.println(str);
}

[Riprickwright]

When I tried it, I got "trim is an unidentified method for the type Entry"

[pbl]

sorry

String str = entries[i].toString().trim();