Question about variable declaration

[macduder83]

All,

I have modified my PHP.ini file so that it reports E_ALL, even the
notices. I wanted to see everything. Question is how do I declare a
variable... bare with me, Im new to PHP.

thx..

[Andy Hassall]

Re: Question about variable declaration

On 21 May 2005 07:42:39 -0700, "macduder83 " <adrian.paveles cu@gmail.com> wrote:
[color=blue]
>I have modified my PHP.ini file so that it reports E_ALL, even the
>notices. I wanted to see everything. Question is how do I declare a
>variable... bare with me, Im new to PHP.[/color]

Variables aren't declared in PHP, at least not separately from initialisation.

Assign a value to the variable before you attempt to read any values from it.

<?php
print $x; // produces a warning
?>

<?php
$x = 1;
print $x; // no warning
?>

--
Andy Hassall / <andy@andyh.co. uk> / <http://www.andyh.co.uk >
<http://www.andyhsoftwa re.co.uk/space> Space: disk usage analysis tool

[macduder83]

Re: Question about variable declaration

Ok... I have the answer. The issue was that the variable was an array.
If I tried to do something like this ::

$myArray[5]; // php screams that it is undeclared, if set E_ALL

So to fix this the first thing you must do is this::

$myArray = NULL; // and then start assigning sizes and whatnot.

So::

$myArray = NULL;
$myArray[sizeof($myOther Array) + 1];

works.

Thanks for the reply...

[Oli Filth]

Re: Question about variable declaration

macduder83 wrote:[color=blue]
> Ok... I have the answer. The issue was that the variable was an[/color]
array.[color=blue]
> If I tried to do something like this ::
>
> $myArray[5]; // php screams that it is undeclared, if set E_ALL
>
> So to fix this the first thing you must do is this::
>
> $myArray = NULL; // and then start assigning sizes and whatnot.
>
> So::
>
> $myArray = NULL;
> $myArray[sizeof($myOther Array) + 1];
>
> works.[/color]

You don't need to declare the size of an array in PHP. It wouldn't make
sense anyway, because an array in PHP is really a hashmap, i.e. the
keys (indices) are arbitrary, and don't have to be sequential, or even
numeric.

e.g.

$a[0] = "blah";
$a[1] = "bleh";
$a[1000] = "blih";
$a["dog"] = "bloh";

$a now has only 4 members, even though one of the keys is 1000.

--
Oli

[Ken Robinson]

Re: Question about variable declaration


macduder83 wrote:[color=blue]
> Ok... I have the answer. The issue was that the variable was an[/color]
array.[color=blue]
> If I tried to do something like this ::
>
> $myArray[5]; // php screams that it is undeclared, if set E_ALL
>
> So to fix this the first thing you must do is this::
>
> $myArray = NULL; // and then start assigning sizes and whatnot.
>
> So::
>
> $myArray = NULL;
> $myArray[sizeof($myOther Array) + 1];[/color]

Actually, you probably want to create a null array:

$myArray = array();

This is helpful if you're using a temporary array in a loop and you
always want to start with an empty array.

Ken

[macduder83]

Re: Question about variable declaration

So thats how you do that. I tried new array(); or something like that
but it complained.