Array manipulation

[Ms.Ranjan]

Hi

i have problem here is the code for it.i have to display the numbers associated with worker id in the following format given below..

String arr[][]=new String[6][2];

arr[0][0]="1573187";
arr[0][1]="CEO id";
arr[1][0]="2871882";
arr[1][1]="worker id";
arr[2][0]="36963";
arr[2][1]="worker id";
arr[3][0]="190444";
arr[3][1]="Manager id";
arr[4][0]="027634";
arr[4][1]="worker id";
arr[5][0]="964956";
arr[5][1]="worker id";


for(int i=0;i<6;i++)
{
   for(int j=i+1;j<6;j++)
     {
        if(arr[i][1].equals(arr[j][1]))

          {
             System.out.println(arr[i][1]+" "+arr[i][0]+" "+arr[j][0]);
            }
       }
  }

ouput:
worker id 2871882 36963
worker id 2871882 027634
worker id 2871882 964956
worker id 36963 027634
worker id 36963 964956
worker id 027634 964956

But
Expected Output is
worker id 2871882 36963 027634 964956

can anyone please help

[Nicholas Jordan]

....can anyone please help

Yes, use a Map of Worker objects keyed by an id field. I enjoy digging through code, but this array indexing is tedious and error prone.

[tim]

Hello Ms.Ranjana

Here is a solution for your problem. A simple way to think about two-dimensional arrays in Java is "there is no two-dimensional array". Yes, it does sound corny, but it's true. Java does not support two-dimensional arrays like Pascal for example. In Java, we have an array of arrays. So in your problem, consider the array to be an array of Strings. So the first index will specify which array you are looking at and the second index will get the element in the current array at the specified index. Consider the table below:

1573187[COLOR="White"]_[/COLOR]CEO id
2871882[COLOR="White"]_[/COLOR]worker id
36963[COLOR="White"]___[/COLOR]worker id

where each row represents an array and each column represents a part of the array. Notice that the second column gives the type of ID and the first gives the value. So the indexes would be

arr[row][column]

Here is the solution:

public class Main{
	public static void main(String[] arg) {
		String arr[][]=new String[6][2];

		arr[0][0]="1573187";
		arr[0][1]="CEO id";
		arr[1][0]="2871882";
		arr[1][1]="worker id";
		arr[2][0]="36963";
		arr[2][1]="worker id";
		arr[3][0]="190444";
		arr[3][1]="Manager id";
		arr[4][0]="027634";
		arr[4][1]="worker id";
		arr[5][0]="964956";
		arr[5][1]="worker id";


		String search = "worker id";
		System.out.print(search);
		for(int i=0;i<6;i++){
			if (arr[i][1].equals(search)){
				System.out.print(" " + arr[i][0]);
			}
		}
		System.out.println();
	}
}

which gives the output

worker id 2871882 36963 027634 964956

I hope this was what you where looking for, Ms.Ranjana. ;)

[Ms.Ranjan]

thankyou....but the problem is i have lots of records like this so i cannot search for worker id alone and display the number related to it

arr[0][0]="1573187";
arr[0][1]="CEO id";
arr[1][0]="2871882";
arr[1][1]="worker id";
arr[2][0]="36963";
arr[2][1]="worker id";
arr[3][0]="190444";
arr[3][1]="Manager id";
arr[4][0]="027634";
arr[4][1]="worker id";
arr[5][0]="964956";
arr[5][1]="CEO id";
arr[6][0]="38914";
arr[6][1]="Manger id";
arr[7][0]="83736";
arr[7][1]="CEO id";

output should be:
CEO id 1573187 964956 83736
worker id 2871882 36963 027634
Manager id 190444 38914

[Norm]

A suggestion on making your code more readable: use a final variable instead of a literal for the indexes.
Instead of arr[i][0] use arr[i][WrkrNbrIdx] & arr[i][WrkrNameIdx]
with these definitions;
final int WrkrNbrIdx = 0;
final int WrkrNameIdx = 1;

System.out.println(arr[i][1]+" "+arr[i][0]+" "+arr[j][0]);

With the above println() statement there is no way to get:

worker id 2871882 36963 027634 964956

I think you need to state what the assignment is more clearly. Looking at you second post, it looks like you want to find all the different worker ids and for each one, output the numbers associated with it.
For this you'll need a way to remember all the unique worker ids, say a Map as Nicholas suggests, using the worker id as the key and storing the numbers as the value associated with the key. So you'll need to think of an class that can store the numbers.

[tim]

Hello Ms.Ranjan

Here is the revision:

public class Main{
	private static String arr[][]=new String[8][2];

	public static void main(String[] arg) {
		arr[0][0]="1573187";
		arr[0][1]="CEO id";
		arr[1][0]="2871882";
		arr[1][1]="worker id";
		arr[2][0]="36963";
		arr[2][1]="worker id";
		arr[3][0]="190444";
		arr[3][1]="Manager id";
		arr[4][0]="027634";
		arr[4][1]="worker id";
		arr[5][0]="964956";
		arr[5][1]="CEO id";
		arr[6][0]="38914";
		arr[6][1]="Man[COLOR="RoyalBlue"]a[/COLOR]ger id";
		arr[7][0]="83736";
		arr[7][1]="CEO id";
		printIDs("CEO id");
		printIDs("worker id");
		printIDs("Manager id");
	}

[COLOR="RoyalBlue"]	public static void printIDs(String key){
		System.out.print(key);
		for(int i = 0; i < arr.length; i++){
			if (arr[i][1].equals(key)){
				System.out.print(" " + arr[i][0]);
			}
		}
		System.out.println();
	}[/COLOR]
}

The code in blue is important. The program outputs:

CEO id 1573187 964956 83736
worker id 2871882 36963 027634
Manager id 190444 38914

Does this help you, Ms.Ranjan? ;)

[Ms.Ranjan]

Thankyou very much Tim,Norm and Nicholas i could solve it...

once again thanks..

[tim]

It's my pleasure Ms.Ranjan ;)

[matt_well]

Hey, this is fun. This is multi-dimentional array.

[tim]

Hey, this is fun. This is multi-dimentional array.

Hello matt_well. I agree with you. You can make a datastructure with any number of dimensions. It only gets more complicated to visualize.

1 dimension[COLOR="White"]__[/COLOR]- a line
2 dimensions - a block
3 dimensions - a cube
4 dimensions - a animated cube
5 dimensions - a line of animated cubes
6 dimensions - a block of animated cubes
7 dimensions - a cube of animated cubes
8 dimensions - a animated cube of animated cubes

This is how I picture them. One can go on forever. ;)