Error when initiating an array using another array in PHP class

Question

I try to initialize an array using another array in php class. Here is the code:

 <?php class test
{
    var $nodeDomain = array
        ("gd88" =>"10.10.104.88", "gd02" =>"10.10.104.2");
    var $node = array
        ("x86-mysql" =>$nodeDomain['gd88'],
         "x86-hbase" =>$nodeDomain['gd02']);

    function show ()
    {
        print_r($node);
    }
}
?>

I got this error: Parse error: syntax error, unexpected T_VARIABLE in /root/workspace/php/array.php on line 6

But when I run the code without using class it works fine. I mean I run the following code:

var $nodeDomain = array
    ("gd88" =>"10.10.104.88", "gd02" =>"10.10.104.2");
var $node = array
    ("x86-mysql" =>$nodeDomain['gd88'],
     "x86-hbase" =>$nodeDomain['gd02']);

I am not quite clear about the difference of php class and php script. Can anyone explain this?

Thanks.

While you're at it: var $foo is deprecated, you must use (private|protected|public) since php 5. Also, print_r($node) won't work because it's in the objects' scope, not in the method scope. Use print_r($this->node) instead. — Berry Langerak
– Berry Langerak, Commented Jul 22, 2011 at 8:23
For further reading, have a look at Properties ... This declaration may include an initialization, but this initialization must be a constant value--that is, it must be able to be evaluated at compile time and must not depend on run-time information in order to be evaluated. — Yoshi
– Yoshi, Commented Jul 22, 2011 at 8:26

Sotomajor · Accepted Answer · 2011-07-22 10:56:50Z

2

You can not use another variables when declaring class members. Try to initialize them in constructor.

<?php class test
{
    var $nodeDomain;
    var $node;

    public function __construct() {
       $this->nodeDomain = array("gd88" =>"10.10.104.88", "gd02" =>"10.10.104.2"); 
       $this->node = array("x86-mysql" =>$this->nodeDomain['gd88'],
         "x86-hbase" =>$this->nodeDomain['gd02']);
    }
    function show ()
    {
        print_r($node);
    }
}
?>

edited Jul 22, 2011 at 10:56

answered Jul 22, 2011 at 8:21

Sotomajor

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1 Comment

Yoshi Over a year ago

Maybe change $nodeDomain to $this->nodeDomain. As $nodeDomain is non-existent in the scope of the constructor.

Waltsu · Accepted Answer · 2011-07-22 08:20:58Z

0

Try put those array-initialazing to the constructor of the test class

answered Jul 22, 2011 at 8:20

Waltsu

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Comments

yankee · Accepted Answer · 2011-07-22 08:24:53Z

0

You just can't reference variables in the field declarations. Where should this variable come from anyway? There are no local variables and no way to position a global statement. (Of course superglobals could work but that's obviously not implemented ;-)) Instead you can do something like this:

<?php class test
{
    var $nodeDomain = array
        ("gd88" =>"10.10.104.88", "gd02" =>"10.10.104.2");
    var $node;

    function __construct()
    {
      $this->node = array
        ("x86-mysql" =>$nodeDomain['gd88'],
         "x86-hbase" =>$nodeDomain['gd02']);
    }
    function show ()
    {
        print_r($node);
    }
}
?>

Beware that $nodeDomain must be in the scope of the constructor somehow. Either it is a global variable, so you need a global $nodeDomain statement before the assignment or you can pass $nodeDomain as constructor argument.

answered Jul 22, 2011 at 8:24

yankee

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1 Comment

arthur bryant Over a year ago

Thanks. I realize why I could not do that. Great help!

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