In my application, a user has_many tickets. Unfortunately, the tickets table does not have a user_id: it has a user_login (it is a legacy database). I am going to change that someday, but for now this change would have too many implications.
So how can I build a "user has_many :tickets" association through the login column?
I tried the following finder_sql, but it does not work.
class User < ActiveRecord::Base
has_many :tickets,
:finder_sql => 'select t.* from tickets t where t.user_login=#{login}'
...
end
I get a weird error:
ArgumentError: /var/lib/gems/1.8/gems/activesupport-2.0.2/lib/active_support/dependencies.rb:402:in `to_constant_name': Anonymous modules have no name to be referenced by
from /var/lib/gems/1.8/gems/activerecord-2.0.2/lib/active_record/base.rb:2355:in `interpolate_sql'
from /var/lib/gems/1.8/gems/activesupport-2.0.2/lib/active_support/dependencies.rb:214:in `qualified_name_for'
from /var/lib/gems/1.8/gems/activesupport-2.0.2/lib/active_support/dependencies.rb:477:in `const_missing'
from (eval):1:in `interpolate_sql'
from /var/lib/gems/1.8/gems/activerecord-2.0.2/lib/active_record/associations/association_proxy.rb:95:in `send'
from /var/lib/gems/1.8/gems/activerecord-2.0.2/lib/active_record/associations/association_proxy.rb:95:in `interpolate_sql'
from /var/lib/gems/1.8/gems/activerecord-2.0.2/lib/active_record/associations/has_many_association.rb:143:in `construct_sql'
from /var/lib/gems/1.8/gems/activerecord-2.0.2/lib/active_record/associations/has_many_association.rb:6:in `initialize'
from /var/lib/gems/1.8/gems/activerecord-2.0.2/lib/active_record/associations.rb:1032:in `new'
from /var/lib/gems/1.8/gems/activerecord-2.0.2/lib/active_record/associations.rb:1032:in `tickets'
from (irb):1
I also tried this finder_sql (with double quotes around the login):
:finder_sql => 'select t.* from tickets t where t.user_login="#{login}"'
But it fails the same way (and anyway, if it worked it would be vulnerable to sql injection).
In a test database, I added a user_id column in the tickets table, and tried this finder_sql:
:finder_sql => 'select t.* from tickets t where t.user_login=#{id}'
Now this works fine. So apparently, my problem has to do with the fact that the users column I am trying to use is a string, not an id.
I searched the net for quite some time... but could not find a clue.
I would love to be able to pass any parameter to the finder_sql, and write things like this:
has_many :tickets_since_subscription,
:finder_sql => ['select t.* from tickets t where t.user_login=?'+
' and t.created_at>=?', '#{login}', '#{subscription_date}']
Edit: I cannot use the :foreign_key parameter of the has_many association because my users table does have an id primary key column, used elsewhere in the application.
Edit#2: apparently I did not read the documentation thoroughly enough: the has_many association can take a :primary_key parameter, to specify which column is the local primary key (default id). Thank you Daniel for opening my eyes! I guess it answers my original question:
has_many tickets, :primary_key="login", :foreign_key="user_login"
But I would still love to know how I can make the has_many :tickets_since_subscription association work.
I think you want the :primary_key
option to has_many
. It allows you to specify the column on the current Table who's value is stored in the :foreign_key
column on the other table.
has_many :tickets, :foreign_key => "user_login", :primary_key => "login"
I found this by reading the has_many docs.
To have something like has_many :tickets_since_subscription you can use named_scopes:
In model add:
named_scope :since_subscription, lambda { |subscription_date| { :conditions => ['created_at > ?', subscription_date] }
With this, you can find what you want like this:
user.tickets.since_subscription 3.days.ago
or
user.tickets.since_subscription user.subscription_date
(of course you need subscription_date column in user model).
You can find more examples here.
If you don't want to use named_scopes you can find what you want with this:
user.tickets.all(:conditions => ['created_at > ?', subscription_date])
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