CODEWITHSUNDEEP
phpclassconstructor
valen
valen

Reputation: 845

attempting to initialize php variables in constructor, getting error

I'm trying to initialize a class attribute within a php constructor method, but am getting the error:

Notice: Undefined variable: _board in C:\wamp\scaleUp\back\objects.php on line 9

code:

<?php
class Board {
public function __construct(){
    for ($x = 9; $x >= 0; $x--) {
        for ($y = 0; $y<10; $y++){
            $row = array();
            $row[$y] = $y;
        }
        $this->$_board = array(); 
            $this->$_board[$x] = $row;
    }
    echo "here";
    echo $this->$board[$x];
}       

 }

 $board =  new Board();

 ?>

Upvotes: -1

Views: 2020

Answers (5)

Mukesh Soni
Mukesh Soni

Reputation: 6668

remove the $ from _board - 

$this->_board = array();

Upvotes: 1

Tabrez Ahmed
Tabrez Ahmed

Reputation: 2960

Here, I have debugged the code for you.

<?php
class Board {
public $_board;
public function __construct(){
    for ($x = 9; $x >= 0; $x--) {
        for ($y = 0; $y<10; $y++){
            $row = array();
            $row[$y] = $y;
        }
        $this->_board = array(); 
            $this->_board[$x] = $row;
    }
    echo "here";
    echo $this->_board[$x+1];/*OR*/print_r($this->_board[$x+1]);
    //$x had to be incremented here.
}       

 }

 $board =  new Board();

 ?>

As others mentioned, you have to follow the syntax: $obj->property, not $obj->$property.

Upvotes: 1

Muhannad A.Alhariri
Muhannad A.Alhariri

Reputation: 3922

You have to define your variable as a member variable suck as 

class object {
 $_board ;
...
...
...
}

and when you want to use it you have to use the following syntax

$this->_board = .....;

I hope this helps you

Upvotes: 0

Ayush
Ayush

Reputation: 42450

It should be

$this->board

You don't need the second $ sign.

Also, in your constructor, in the inner loop, you are re-initializing $row as an array in every iteration. Is that intended?

Upvotes: 0

ThiefMaster
ThiefMaster

Reputation: 318748

The syntax to access an object field is $obj->field, not $obj->$field (unless you want to access the field name that is stored in $field).

Upvotes: 2

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