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Home/ Questions/Q 877059
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Editorial Team
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Editorial Team
Asked: May 15, 20262026-05-15T11:32:36+00:00 2026-05-15T11:32:36+00:00

I am building a Rails application accessing a legacy system. The data model contains

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I am building a Rails application accessing a legacy system. The data model contains Customers which can have one or more Subscriptions. A Subscription always belong to one and only one Customer. Though not needed, this association is represented through a join table “subscribes”, which do not have an id column:

 Column          |  Type   | Modifiers
-----------------+---------+-----------
 customer_id     | integer | not null
 subscription_id | integer | not null

I have this coded as a has_and_belongs_to_many declarations in both Customer and Subscription

class Customer < Activerecord::Base
  has_and_belongs_to_many :subscriptions, :join_table => "subscribes",
    :foreign_key => "customer_id", :association_foreign_key => "subscription_id"
end
class Subscription < Activerecord::Base
  has_and_belongs_to_many :customers, :join_table => "subscribes",
    :foreign_key => "subscription_id", :association_foreign_key => "customer_id"
end

The problem I have is that there can only ever be one customer for each subscription, not many, and the join table will always contain at most one row with a certain customer_id.
And thus, I don’t want the association “customers” on a Subscription which returns an array of (at most one) Customer, I really do want the relation “customer” which returns the Customer associated.

Is there any way to force ActiveRecord to make this a 1-to-N relation even though the join table itself seems to make it an N-to-M relation?

–Thomas

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  1. Editorial Team
    Editorial Team
    2026-05-15T11:32:36+00:00Added an answer on May 15, 2026 at 11:32 am

    I don’t know if you can do that with ActiveRecord, but if you want any solution to this problem, you can define a customer method on subscription model:

    class Subscription < Activerecord::Base
       has_and_belongs_to_many :customers, :join_table => "subscribes",
          :foreign_key => "subscription_id",
          :association_foreign_key => "customer_id"
    
       def customer
          self.customers.first
       end
    end
    
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