Mongoid嵌入一個類作爲多個字段

Question

我試圖跟蹤用戶的當前，先前和選擇（期末更改）訂閱狀態。

在月結束時，如果chosen_subscription != current_subscription，current_subscription被更改。

class User 
    include Mongoid::Document 

    embeds_one :previous_subscription, class_name: "Subscription" 
    embeds_one :current_subscription, class_name: "Subscription" 
    embeds_one :chosen_subscription, class_name: "Subscription" 
end 

class Subscription 
    include Mongoid::Document 

    embedded_in :user 

    field :plan, type: String 
    field :credits, type: Integer 
    field :price_per_credit, type: BigDecimal 

    field :start, type: Date 
    field :end, type: Date 
end 

Mongoid要我來指定這個更多的方式，沒有任何意義對我說：

Mongoid::Errors::AmbiguousRelationship: 

Problem: 
    Ambiguous relations :previous_subscription, :current_subscription, :chosen_subscription defined on User. 
Summary: 
    When Mongoid attempts to set an inverse document of a relation in memory, it needs to know which relation it belongs to. When setting :user, Mongoid looked on the class Subscription for a matching relation, but multiples were found that could potentially match: :previous_subscription, :current_subscription, :chosen_subscription. 
Resolution: 
    On the :user relation on Subscription you must add an :inverse_of option to specify the exact relationship on User that is the opposite of :user. 

這發生在我覆蓋現有current_subscription。我想，在這個時候，Mongoid想要取消註冊舊訂閱用戶的訂閱。

當然，每個認購對象只屬於一個用戶和user == user.previous_subscription.user == user.current_subscription.user == user.chosen_subscription.user

然而，它並不意義，我要告訴Subscription即user是三個中的任何一個的倒數。

我應該如何構建它？

Answer 1

正如它所說：您必須添加：inverse_of選項。試試這個：

class User 
    include Mongoid::Document 

    embeds_one :previous_subscription, class_name: "Subscription", inverse_of: :previous_subscription 
    embeds_one :current_subscription, class_name: "Subscription", inverse_of: :current_subscription 
    embeds_one :chosen_subscription, class_name: "Subscription", inverse_of: :chosen_subscription 
end 

class Subscription 
    include Mongoid::Document 

    embedded_in :user, inverse_of: :previous_subscription 
    embedded_in :user, inverse_of: :current_subscription 
    embedded_in :user, inverse_of: :chosen_subscription 

    field :plan, type: String 
    field :credits, type: Integer 
    field :price_per_credit, type: BigDecimal 

    field :start, type: Date 
    field :end, type: Date 
end 

Answer 2

每個mongoid關係都有一個反比關係，用來正確設置兩邊。大多數情況下，它可以根據命名約定，使用的類等正確識別，但有時您需要明確定義它們。

問題mongoid抱怨這裏的是，在User模型中定義的關係是模糊的，因爲mongoid不能確定它們中的哪映射到Subscriptionsubscription.user關係。此外，當使用subscription.user = some_user時，mongoid無法識別您是否需要在訂閱對象的用戶中設置previous_subscription，current_subscription或chosen_subscription。即使你沒有明確地調用這個方法，當你設置一個關係時，你可以面對這些問題，因爲mongoid試圖設置反向。

我不確定這是否是構建系統的正確方法，但您可以在關係中定義inverse_of: nil以告訴mongoid在設置關係時不設置反向關係。只有在您的代碼中不需要使用subscription.user=時才應該使用此方法。 subscription.user也可能有問題，但如果您確實需要它們，則可以使用未公開的方法doc._parent或doc._root。這種方法雖然解決了問題，但有其自身的缺點。我試圖將它們與下面的代碼顯示在這裏：

class Tester 
    include Mongoid::Document 

    field :name, type: String 

    embeds_one :first_module, class_name: 'TestModule', inverse_of: nil 
    embeds_one :last_module, class_name: 'TestModule', inverse_of: nil 

end 

class TestModule 
    include Mongoid::Document 

    field :name, type: String 

    # not needed as such, but required to tell mongoid that this is a embedded document 
    embedded_in :tester, inverse_of: nil 
end 

t1 = Tester.new(name: 't1') 
m1 = t1.build_first_module(name: 't1m1') 
m2 = t1.build_last_module(name: 't1m2') 

t1.first_module == m1 #=> true 
t1.last_module == m2 #=> true 

m1.tester    #=> nil 
m2.tester    #=> nil 

t1.save 
t1 = Tester.last 
m1 = t1.first_module 
m2 = t1.last_module 

t1.first_module == m1 #=> true 
t1.last_module == m2 #=> true 

m1.tester    #=> nil 
m2.tester    #=> nil 

已經有幾對這樣的事情上mongoid問題跟蹤過一個討論：#1677 & #3086