$foo->bar = 42;

Whether bar is defined or not, it will work, last I heard.

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----- Original Message -----
From: Bret R. Zaun <[EMAIL PROTECTED]>
Newsgroups: php.general
Sent: Tuesday, August 14, 2001 10:50 AM
Subject: How dynamically create objects

> Hello,
> I would like to create an object from a db result like MySQL does in
> mysql_fetch_object
> How can I create an object with members which names are set at runtime ?
> Regards
> Bret

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