Victor Kapustin wrote: > > echo $snippet->code; > > displays some part of snippet and it depends what exactly is in it. > > You should have all snippet code in page source. > > it DOES NOT. > > > And &(snippet.code:p); > > displays snippet's name....... > > And even &(snippet.code); displays the code. That's what worries.
If &(snippet.code); displays the code, than it must be present in the loaded object. If you want the code to be executed, you have to do &(snippet.code:p); Emile --------------------------------------------------------------------- To unsubscribe, e-mail: [EMAIL PROTECTED] For additional commands, e-mail: [EMAIL PROTECTED]
