Re: DB_DataObject_FormBuilder: where to set select_display_field?

On Mon, 07 Mar 2005 01:33:18 +0100, Marek Kilimajer <lists@kilimajer.net> wrote: > Dot Deb wrote: > > I'm quite new to PEAR, in general, and far more new to > > DB_DataObject_FormBuilder. > > The main problem is that I can't find reliable documentation. > > So, the main objective of this message is to get some good references. > > > > Looking at the the particular problem: I can't manage to have "hobby > > name" in a select > > for a "person" form, referencing to a (secondary) "hobby" table. > > The said "Select" shows only hobby_ids. > > How can I to set the "select_display_field" option? > > Put this in your DataObjects_Hobby class definition: > > var $fb_linkDisplayFields = array('hobby_name'); > > (under ###END_AUTOCODE) > > > I read the "rapidprototyping.pdf" presentation but I can't obtain the > > results shown there. > > > > It's obsolete, the new one is 404 Not Found > Yes, it is very obsolete. I'll tell Markus about the 404. If you look in that page, there's formbuilder docs there. I keep those up to date. Just look at the wiki pages there for more. -- Justin Patrin