Sorting on a related STI-field

Rails
1

I have the following models:
Event
name
location
customer_id
with relation
belongs_to :customer

Person
name
address
city
type

Customer < Person
has_many :events

Now I want to have a list, sorted on the name of the customer.

This doesn’t work
Event.find(:all, :include => :customer, :order => ‘customer.name’)

Or isn’t this possible at all, just back to plain sql?

Greetz, Adrie

2

sure you can write in plain sql using find_by_sql method. but it is an
easy query and it would easier doing it directly in rails code.
I’m not sure i understand what you want maybe it is :
Costumer.find :all, :include => :events, :order => ‘customers.name’

and a question: maybe you made an errors by typing but why you wrote
customer (singular) in ‘customer.name’ ?

On Jan 24, 1:31 am, Adrie D. [email protected]

3

Jean-sébastien Jney wrote:

Costumer.find :all, :include => :events, :order => ‘customers.name’

Jean,

I want the Events (a selection on event.starttime) listed, but sorted
bij de Customer’s name. Not all Events have a customer, so your query
went give them.

That why i tried:
Event.find(:all, :include => :customer, :order => ‘customer.name’)

Anyway, thanks for your reply
Adrie

4

On 24 Jan 2008, at 12:48, Adrie D. wrote:

That why i tried:
Event.find(:all, :include => :customer, :order => ‘customer.name’)

just add a condition: ‘customers.id IS NOT NULL’ (for mysql, other dbs
will have something equivalent’).
There’s also :joins instead of :include (which does mostly the same
joining, but doesn’t actually instantiate the customer objects)

Fred

5

Frederick C. wrote:

just add a condition: ‘customers.id IS NOT NULL’ (for mysql, other dbs
will have something equivalent’).
There’s also :joins instead of :include (which does mostly the same
joining, but doesn’t actually instantiate the customer objects)

Fred

Fred,

I worked it out, this is the way:

Event.find(:all, :include => :customer, :order => ‘people.name’)

The NULL clause doesn’t seem necessary, just use the plural name of the
table, not the model.

Thanks for the support
Adrie