Re: Object reference into variable?

I strongly suggest you put some pressure on you web provider to at least upgrade it to the 4.4.7, as 4.2 is way too old to be secure :) Unfortunately I don't know how this could be accomplished in PHP4, but a quick google query returned these: http://www.hat.net/geeky/php_tricks_-_php_5_clone_in_php4 http://www.dgx.cz/trine/item/how-to-emulate-php5-object-model-in-php4 Someone with more experience might shed some light into this.. Regards, Hamza. ""Sándor Tamás (GMail)"" <santasoft@gmail.com> wrote in message news:008501c7d8d2$a64f5fe0$3c646464@stgepe... 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)"" <santasoft@gmail.com> wrote in message news:003c01c7d8cd$8a84e910$3c646464@stgepe... > 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