Iterator interface problem in an ArrayList

**Ash maker** · 09-02-2010

Hello everyone,
I have a problem when using the iterator interface in an ArrayList. Here is my code

Code:

Public class Person {
	private String name; 
Public Person (string name) {
this.name = Name;
}
Public String getName() {
 return name;
}

I need a help with this, if any one knows how to do this, then please help me.

**SoftWore** · 09-02-2010

Hello,
I think to get the desired output you must overload the toString person in your class

Code:

Public String toString() {
  return name;
}

Just do this and see if the program runs, if not then you can try some other alternative, but till now I do not know one. I will search for it and when I will know about it, I will post back with it.

**InterNetWorKed** · 09-02-2010

Hello,
Even I think you must overload the toString method of your class person. Why? Attribute name is a String, it may well use its getter for display in a System.out.println. Otherwise your code looks good! (except for errors in the code posted: typos probably ) Only thing you can do:

Code:

for (Iterator <person> i = lsper.iterator (); i.hasNext ())

This is why it is necessary. Just include this in your code and see if it works, by my knowledge it should work.

**NetWorkInG** · 09-02-2010

Hello,
Your code for typos almost works. As you use parameterized types (generics) I concluded that you are using Java 1.5 or higher. You can then use the extended loop that I find more convenient than iterators:

Code:

for (Person Person: lsper) {
  System.out.System.out.println(person.getName());
}

I hope that you have included this in your code and the program runs successfully.

**GlassFish** · 09-02-2010

Hello,
I have a sample code with me , I have posted it, see below. If this is what you are find then well and good

Code:

public class tic
{
private int nmln;
private ArrayList ltln;

public tic()
{
ltln = new ArrayList();
nmln = 1;
}
public void showNumbers()
{
Iterator it = ltln.iterator();
while(it.hasNext()){
Numbers ltln = (Numbers) it.next();
System.out.println(ltln.toString());
}
}

}

**New ID** · 09-02-2010

Hello,
Even I was find a solution for this problem but unfortunately I could not find one. If you have any solution or any idea about it then please help me with posting. Here is my code

Code:

Public class PersonneTest {
	Public static void hand(String[] args) {
			
	ArrayList <ps> lsp = new ArrayList <ps>();

lsp.add(new Person("sm"));
lsp.add(new Person("sj"));
		
		for (Iterator i = lsp.iterator()i.hasNext();)
		{
		Person ps = (Person)i.next();
System.out.System.out.println(ps.getName());
		}
	}