Re: Object reference into variable?

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)"" <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