如何在Rails中爲單個表繼承建模人員和公司？

Question

在我的進銷存應用程序invoices可以發送到company或person。據我瞭解，這是Rails單表繼承（STI）的一個很好的用例。由於這兩種類型有許多共同的屬性和功能，我計算過，一個超一流的Recipient可能是一個很好的路要走：

class Recipient < ActiveRecord::Base 
end 

class Company < Recipient 
    has_many :people 
end 

class Person < Recipient 
    belongs_to :company 
end 

我也明白，我需要在Recipient模型的屬性type。

困擾我的唯一的事實是person可能（或可能不）屬於company。這怎麼可以在Rails中模擬？通常，我只需將另一個數據庫字段company_id添加到people表中。但是這裏只有一個表格（recipients）。那麼這怎麼做呢？

感謝您的任何幫助。

Answer 1

的結構看起來是這樣的：

class Recipient < ActiveRecord::Base 
    has_many :invoices 
end 

class Company < Recipient 
    has_many :people 
end 

class Person < Recipient 
    belongs_to :company 
end 

class Invoice < ActiveRecord::Base 
    belongs_to :recipients 
end 

# Schema 
create_table "invoices", force: :cascade do |t| 
    t.integer "recipient_id" 
    t.datetime "created_at", null: false 
    t.datetime "updated_at", null: false 
    t.index ["recipient_id"], name: "index_invoices_on_recipient_id" 
end 

create_table "recipients", force: :cascade do |t| 
    t.integer "company_id" 
    t.string "type" 
    t.datetime "created_at", null: false 
    t.datetime "updated_at", null: false 
end 

我在控制檯只是嘗試：

> Recipient.all 
=> [#<Company:0x007fd55d797220 
    id: 1, 
    company_id: nil, 
    type: "Company", 
    created_at: Fri, 04 Aug 2017 10:57:41 UTC +00:00, 
    updated_at: Fri, 04 Aug 2017 10:57:41 UTC +00:00>, 
#<Person:0x007fd55d796730 
    id: 2, 
    company_id: 1, 
    type: "Person", 
    created_at: Fri, 04 Aug 2017 10:57:41 UTC +00:00, 
    updated_at: Fri, 04 Aug 2017 10:57:41 UTC +00:00>, 
#<Person:0x007fd55d796208 
    id: 3, 
    company_id: nil, 
    type: "Person", 
    created_at: Fri, 04 Aug 2017 10:57:41 UTC +00:00, 
    updated_at: Fri, 04 Aug 2017 10:57:41 UTC +00:00>] 

> Person.last.company 
    Person Load (0.2ms) SELECT "recipients".* FROM "recipients" WHERE "recipients"."type" IN ('Person') ORDER BY "recipients"."id" DESC LIMIT ? [["LIMIT", 1]] 
=> nil 

> Person.first.company 
    Person Load (0.2ms) SELECT "recipients".* FROM "recipients" WHERE "recipients"."type" IN ('Person') ORDER BY "recipients"."id" ASC LIMIT ? [["LIMIT", 1]] 
    Company Load (0.2ms) SELECT "recipients".* FROM "recipients" WHERE "recipients"."type" IN ('Company') AND "recipients"."id" = ? LIMIT ? [["id", 1], ["LIMIT", 1]] 
=> #<Company id: 1, company_id: nil, type: "Company", created_at: "2017-08-04 10:57:41", updated_at: "2017-08-04 10:57:41"> 

Answer 2

class Recipient < ActiveRecord::Base 
end 

class Company < Recipient 
    has_many :people, class_name: "Recipient", foreign_key: 'parent_id' 
end 

class Person < Recipient 
    belongs_to :company, class_name: "Recipient", foreign_key: 'parent_id' 
end 

只需將parent_id添加到收件人遷移。 這就是它簡單和快速，你得到你想要的一個模型兩個STI和has_many和belongs_to之間company和person。