Sorry.. 1 more thing.
php5 does not use var.
use public $variable=value; instead.
public is only within a class however. you cannot use it outside.
On Thursday 20 October 2005 09:35 pm, Bob Hartung wrote:
> Hi all,
> I'm trying to get started in OOP with PHP. I have the following
> short code snipped. I'f I comment out the 'class Test' definition and
> the following references to it, it prints
> This is outside the php code block
>
> and
>
> Start defining the class here:
>
> If I do not comment out the code as noted, then the page returned is
> totally blank. It does this with or without using the constructor.
> PHP 5 on apache. Same behavior both on Win32 and FC4.
>
> All help appreciated to get me going.
>
> Code Snippet:
>
> <html>
>
> <head>
> <title>PHP Class testing</title>
>
> </head>
> <body>
> <br>
> <P>This is outside the php code block</P>
> <br>
> <?php
> echo "Start defining the class here: <BR>" ;
> /* class Test
> {
>
> function __constructor()
> {
> var $saying ;
> $saying = "Im in the Test Class" ;
> }
>
> function get()
> {
> return $saying ;
>
> }
>
> var $liveclass ;
> $liveclass = new Test ;
> echo $liveclass->get() ;
> echo "<BR>" ;
> echo "This is in the php code block" ;
> */
> ?>
>
> </body>
> </html>
--
PHP General Mailing List (http://www.php.net/)
To unsubscribe, visit: http://www.php.net/unsub.php
