Apache Fatal Error Problem

[wilz04]

Hello friends,

I have problems with this:

MenuItem.class.php:

<?php
	class MenuItem {
		
		var $label;
		var $href;
		var $target;
		
		function MenuItem($label="", $href="", $target="") {
			$this->label = $label;
			$this->href = $href;
			$this->target = $target;
		}
		
		function isEmpty() {
			return $this->label == "" && $this->href == "" && $this->target == "";
		}
		
		function toString() {
			return "<td><a href='".$this->href."' target='".$this->target."'>".$this->label."</a></td>";
		}
		
	}
?>

MenuBar.class.php:

<?php
include_once("MenuItem.class.php");

class MenuBar {
	
	var $items;
	
	function MenuBar() {
		$this->items = array(new MenuItem());
	}
	
	function add($item) {
		if ($this->items[0]->isEmpty()) {
			$this->items[0] = $item;
		} else {
			$this->items[count($this->items)] = $item;
		}
	}
	
	function show() {
		$me = "<table><tr>";
		$j = count($this->items);
		for ($i=0; $i<$j; $i++) {
			$me .= $this->items[$i]->toString();
		}
		$me .= "</tr></table>";
	}
	
}
?>

menu.php:

<?php
	include_once("../scripts/MenuBar.class.php");
	
	$menu = new MenuBar();
	$menu->add("Google", "http://www.google.co.cr/", "_blank");
	$menu->show();
?>

The Apache return the follow error:

Fatal error: Call to a member function on a non-object in /home/iconco/public_html/invent/scripts/MenuBar.class.php on line 25

Can Helpme??

[PeejAvery]

Welcome to the forums wilz04. Please do not post over someone else's problem. In fact, you posted on a thread 2 years old. I will have a moderator split this for you.

Second, please use [code] tags when posting code. It makes things easier to read. You would use them in the following manner.
[code]
your code here
[/code]

As concerning your problem, the log has given you some help. What is the 25th line of MenuBar.class.php?

[wilz04]

Yes, I Know.. but the I don't resolve the error.

I declared all the vars (instances) and call my functions correctly, but the apache continue says: Call to a member function on a non-object!!!!, but it 's a object!!!
I declare the items array as var of the class MenuBar, and the following code is the body of the her 'show' function, and I assign an array in the constructor of this class.
In the function 'add', I assign to '$items[0]' the value of '$item', param of the function, then '$items' is not null.
And 'toString()' is a function of class 'MenuItem', type of '$items', he returns a string.
I belive that all is good, BUT NOT!!

MenuBar.class.php (only 'show' function):

function show() {
	$me = "<table><tr>";
	$j = count($this->items);
	for ($i=0; $i<$j; $i++) {
		$me .= $this->items[$i]->toString();//line of error (25)
	}
	$me .= "</tr></table>";
	echo $me;
}

I realy need help!!

[PeejAvery]

Well, where is your function toString()? That is a JavaScript function, but in PHP it does not exist. In fact, variables act differently.

[wilz04]

This is my function 'toString()'.
She is in the class 'MenuItem'
the vars '$href', '$target', and '$label was declared as members of this class, and in her constructor I initialized this vars with params values.

MenuItem.class.php (only 'toString()' function):

function toString() {
	return "<td><a href='".$this->href."' target='".$this->target."'>".$this->label."</a></td>";
}

Help me PeejAvery!!

Note: Can you write in spanish?

[PeejAvery]

In your code, $this->items[$i] must be returning a non-object. Have you echoed them to see what they are obtaining?

[wilz04]

I echoed '$this->items[$i]' to see what they are obtaining, and show 'Google'.

[PeejAvery]

Well, as you can see Google is just the first of three parts of an object/array. Try changing the show function to the following code.

Como ya puedes ver, Google es solamente la primera de tres partes de un array. Prueba el cambio en la funcion show.

function show() {
    $me = "<table><tr>";
    $me .= $this->items->toString();
    $me .= "</tr></table>";
    echo $me;
} 

EDIT: Por que quieres que escriba yo en el castellano?

[wilz04]

Thanks PeejAvery, I see the error, it's in the menu.php:

<?php
	include_once("../scripts/MenuBar.class.php");
	
	$menu = new MenuBar();
	$menu->add(new MenuItem("Google", "http://www.google.co.cr/", "_blank"));//line (5) of error (resolved)
	$menu->show();
?>

previously I send three params at the function 'add()' of class 'MenuItem'
Before:

$menu->add("Google", "http://www.google.co.cr/", "_blank"));//line 5, bad!

Now, I send only one param, yes, my 'add()' only recv one param (type of 'MenuItem')!!
After:

$menu->add(new MenuItem("Google", "http://www.google.co.cr/", "_blank"));//line 5, good!

Thanks again PeejAvery!!

[PeejAvery]

Glad you solved it. So why did you want me to write in Spanish?

[wilz04]

Gracias PeejAvery, Si, logre solucionar el problema, pero me ayudaste muchisimo!!