Hi, As per another member's suggestion, I'm trying to user STI. So I have this table: CREATE TABLE `settings` ( `id` int(11) unsigned NOT NULL auto_increment, `type` char(40) NOT NULL default 'General', `name` char(40) NOT NULL default '', `value` char(128) NOT NULL default '', `updated_on` datetime default NULL, `comments` char(255) default NULL, PRIMARY KEY (`id`) ) ENGINE=MyISAM DEFAULT CHARSET=utf8; And a row with a type = 'General', and name = 'front-page' Then I have a setting.rb model, and also general_setting.rb and user_setting.rb, where I only have: class <TYPE>Setting < Setting end So, if I do General.find_by_name('front-page'), I expected to get the value of the corresponding row, instead I get a nil object. What could I possibly be doing wrong? Thanks! Ivan V.
on 2006-01-07 00:01
on 2006-01-07 02:01
[Note: HowTo in the subject suggests you're providing a HowTo. You're not. You're asking a question.] On Fri, Jan 06, 2006 at 03:59:18PM -0600, IvÃ¡n Vega R. wrote: > PRIMARY KEY (`id`) > So, if I do General.find_by_name('front-page'), I expected to get the > value of the corresponding row, instead I get a nil object. The value of the type field should be the name of the subclass you wish to instantiate for that row. So if your class is named GeneralSetting, you should be inserting type='GeneralSetting'. - Matt
on 2006-01-07 12:08
Iván The 'type' column is used by Rails in the background and, most of the time, there is no need for you to manipulate it explicitly. So, if you have class Foo < ActiveRecord and class SuperFoo < Foo , you can query them directly: allRecords = Foo.find(:all) onlyTheSuperFoos = SuperFoo.find(:all) Did that answer your question? Alain
on 2006-01-07 19:42
Matthew, although a bit crankyly ;-), had the answer. I wasn't using the correct type name for each class. Although you do make a point. I should've used your method to test how STI worked in the first place. Thanks! Ivan V.