Re: DB_DataObject loading linked values

On 19 Aug 2004 23:08:39 -0000, Daniel Kopp <until-june-04@daniu.de> wrote: > Hi! > > Justin Patrin wrote: > > > Why do you need all of the linked values? A Form perhaps? Have you > > looked into DB_DataObject_FormBuilder? > > Well - it's a bit more difficult ... > FormBuilder can do a lot. ;-) > > That said, I'm sure you don't need eval(). Take this snip of code > > for example: $var = 'someVarName'; > > $o = new Obj(); > > $o->someVarName = 'hello'; > > echo $o->$var; > > > > This will echo "hello". > > I just found out about this also ... And probably I will do that - but > is there also a way to find out which are the linked values? Then I > would not need a predefined array ... If you want which fields are links: $do = DB_DataObject::factory('tableName'); $links = $do->links(); $linkFields = array_keys($do->links); > > regards > Daniel > > -- > PEAR General Mailing List (http://pear.php.net/) > To unsubscribe, visit: http://www.php.net/unsub.php > > > !DSPAM:4124d0c420573569737329! > > -- DB_DataObject_FormBuilder - The database at your fingertips http://pear.php.net/package/DB_DataObject_FormBuilder paperCrane --Justin Patrin--