Have you asked your web provider if they have anything newer than 4.2? I
had to ask my profider if they could upgrade my PHP and they already had, I
just had to set my PHP to be handled by PHP5 in my Apache handlers.
Also, if they won't/can't upgrade that's a bad sign. Move away quickly.
- Dan
""Sándor Tamás (GMail)"" <[EMAIL PROTECTED]> wrote in message
news:[EMAIL PROTECTED]
I forgot to tell you that I have to use PHP4.2 because of my web provider.
Anything besides of _clone?
SanTa
----- Original Message -----
From: Hamza Saglam
To: "Sándor Tamás (GMail)"
Sent: Tuesday, August 07, 2007 10:49 AM
Subject: Re: Object reference into variable?
You may find the following article interesting. It talks about object
cloning, which I think what you are after...
http://www.phpfreaks.com/phpmanual/page/language.oop5.cloning.html
Regards,
Hamza.
""Sándor Tamás (GMail)"" <[EMAIL PROTECTED]> wrote in message
news:[EMAIL PROTECTED]
> Hello,
>
> I have a question (what a surprise :-) )
>
> I browsed the archives, but didn't find what I'm looking for.
> I have an object instance stored in some variable (like $myobject). I
want
> to save this instance into another variable (like $tempobject). I
thought,
> this will be the way:
> $myobject = new MyObject();
> $tempobject = $myobject;
> (or I tried this too:
> $tempobject = &$myobject)
>
> None of them worked, like I cannot access methods in the object:
> $tempobject->Foo()
> gives me an error: calling to method on a non-object variable.
>
> How can I tell PHP that $tempobject is a class instance of MyObject, and
> that I can use it's methods?
>
> Thanks,
> SanTa
--
PHP General Mailing List (http://www.php.net/)
To unsubscribe, visit: http://www.php.net/unsub.php
