Re: DB_DataObject_FormBuilder: where to set select_display_field?

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