|  | Posted by M. Trausch on 11/17/05 06:18 
sam wrote:>
 > $db = new $class_name ($user, $pwd, $dbName);
 >
 > Now, PHP knows that $class_name is a variable that contains the name of the
 > class.  That why it should work.
 >
 
 Hrm.
 
 I didn't realize that you could use a variable like that in PHP.  That's
 a pretty nifty feature, actually.  I never tried it quite like that,
 probably because I'm thinking too much like I'm working in other
 languages.  What I wound up doing was working with eval(); and taking
 advantage of PHP's variable substitution within double-quoted strings to
 achieve the same effect, which worked rather nicely.  I'll have to keep
 this in mind for the future, though.  That's pretty cool.
 
 Thanks,
 Mike
 
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