has_one/has_many rails association with alternate source ID other then primary key












3














I have 3 models.



Model A belongs to model B and model B has many model C



by using through I can set in model A class has_many :c,through: :b



Now whenever I call a_instance.cs It would JOIN with table B unnecessarily What I want is direct association of A & C using b_id in both A & C class.



So how can I write a has_one/has_many rails association without through clause when both Source and Destination entities has same foreign_key of 3rd entity? (b_id for this example)



class C
belongs_to :b #b_id column is there in DB
end

class B
has_many :cs
end

class A
belongs_to :b #b_id column is there in DB
has_many :cs , through: :b #This is the association in Question

#WHAT I WANT TO DO. So something in place of primary_key which would fetch C class directly.
has_many :cs ,class_name: 'C',foreign_key: 'b_id',primary_key: 'b_id'
end









share|improve this question


















  • 1




    Why is this tagged as ruby-on-rails-4 and ruby-on-rails-5? The solution probably doesn't make a difference in this case, but it's advisable to actually specify the version you're using.
    – Tom Lord
    Oct 22 '18 at 8:20








  • 4




    The last has_many with foreign_key and primary_key works for me, what issues are you having? Specifying class_name is not necessary, tested with Rails 5.1.4.
    – Marcin Kołodziej
    Nov 15 '18 at 18:29






  • 1




    Agree, the last has_many works for me and also works w/o the primary_key.
    – Derek
    Nov 20 '18 at 4:22
















3














I have 3 models.



Model A belongs to model B and model B has many model C



by using through I can set in model A class has_many :c,through: :b



Now whenever I call a_instance.cs It would JOIN with table B unnecessarily What I want is direct association of A & C using b_id in both A & C class.



So how can I write a has_one/has_many rails association without through clause when both Source and Destination entities has same foreign_key of 3rd entity? (b_id for this example)



class C
belongs_to :b #b_id column is there in DB
end

class B
has_many :cs
end

class A
belongs_to :b #b_id column is there in DB
has_many :cs , through: :b #This is the association in Question

#WHAT I WANT TO DO. So something in place of primary_key which would fetch C class directly.
has_many :cs ,class_name: 'C',foreign_key: 'b_id',primary_key: 'b_id'
end









share|improve this question


















  • 1




    Why is this tagged as ruby-on-rails-4 and ruby-on-rails-5? The solution probably doesn't make a difference in this case, but it's advisable to actually specify the version you're using.
    – Tom Lord
    Oct 22 '18 at 8:20








  • 4




    The last has_many with foreign_key and primary_key works for me, what issues are you having? Specifying class_name is not necessary, tested with Rails 5.1.4.
    – Marcin Kołodziej
    Nov 15 '18 at 18:29






  • 1




    Agree, the last has_many works for me and also works w/o the primary_key.
    – Derek
    Nov 20 '18 at 4:22














3












3








3







I have 3 models.



Model A belongs to model B and model B has many model C



by using through I can set in model A class has_many :c,through: :b



Now whenever I call a_instance.cs It would JOIN with table B unnecessarily What I want is direct association of A & C using b_id in both A & C class.



So how can I write a has_one/has_many rails association without through clause when both Source and Destination entities has same foreign_key of 3rd entity? (b_id for this example)



class C
belongs_to :b #b_id column is there in DB
end

class B
has_many :cs
end

class A
belongs_to :b #b_id column is there in DB
has_many :cs , through: :b #This is the association in Question

#WHAT I WANT TO DO. So something in place of primary_key which would fetch C class directly.
has_many :cs ,class_name: 'C',foreign_key: 'b_id',primary_key: 'b_id'
end









share|improve this question













I have 3 models.



Model A belongs to model B and model B has many model C



by using through I can set in model A class has_many :c,through: :b



Now whenever I call a_instance.cs It would JOIN with table B unnecessarily What I want is direct association of A & C using b_id in both A & C class.



So how can I write a has_one/has_many rails association without through clause when both Source and Destination entities has same foreign_key of 3rd entity? (b_id for this example)



class C
belongs_to :b #b_id column is there in DB
end

class B
has_many :cs
end

class A
belongs_to :b #b_id column is there in DB
has_many :cs , through: :b #This is the association in Question

#WHAT I WANT TO DO. So something in place of primary_key which would fetch C class directly.
has_many :cs ,class_name: 'C',foreign_key: 'b_id',primary_key: 'b_id'
end






ruby-on-rails ruby ruby-on-rails-4 activerecord ruby-on-rails-5






share|improve this question













share|improve this question











share|improve this question




share|improve this question










asked Oct 22 '18 at 8:09









Qaisar Nadeem

1,815718




1,815718








  • 1




    Why is this tagged as ruby-on-rails-4 and ruby-on-rails-5? The solution probably doesn't make a difference in this case, but it's advisable to actually specify the version you're using.
    – Tom Lord
    Oct 22 '18 at 8:20








  • 4




    The last has_many with foreign_key and primary_key works for me, what issues are you having? Specifying class_name is not necessary, tested with Rails 5.1.4.
    – Marcin Kołodziej
    Nov 15 '18 at 18:29






  • 1




    Agree, the last has_many works for me and also works w/o the primary_key.
    – Derek
    Nov 20 '18 at 4:22














  • 1




    Why is this tagged as ruby-on-rails-4 and ruby-on-rails-5? The solution probably doesn't make a difference in this case, but it's advisable to actually specify the version you're using.
    – Tom Lord
    Oct 22 '18 at 8:20








  • 4




    The last has_many with foreign_key and primary_key works for me, what issues are you having? Specifying class_name is not necessary, tested with Rails 5.1.4.
    – Marcin Kołodziej
    Nov 15 '18 at 18:29






  • 1




    Agree, the last has_many works for me and also works w/o the primary_key.
    – Derek
    Nov 20 '18 at 4:22








1




1




Why is this tagged as ruby-on-rails-4 and ruby-on-rails-5? The solution probably doesn't make a difference in this case, but it's advisable to actually specify the version you're using.
– Tom Lord
Oct 22 '18 at 8:20






Why is this tagged as ruby-on-rails-4 and ruby-on-rails-5? The solution probably doesn't make a difference in this case, but it's advisable to actually specify the version you're using.
– Tom Lord
Oct 22 '18 at 8:20






4




4




The last has_many with foreign_key and primary_key works for me, what issues are you having? Specifying class_name is not necessary, tested with Rails 5.1.4.
– Marcin Kołodziej
Nov 15 '18 at 18:29




The last has_many with foreign_key and primary_key works for me, what issues are you having? Specifying class_name is not necessary, tested with Rails 5.1.4.
– Marcin Kołodziej
Nov 15 '18 at 18:29




1




1




Agree, the last has_many works for me and also works w/o the primary_key.
– Derek
Nov 20 '18 at 4:22




Agree, the last has_many works for me and also works w/o the primary_key.
– Derek
Nov 20 '18 at 4:22












2 Answers
2






active

oldest

votes


















1














An example, Teacher and Student models belong to Group, Teacher has many Student:



class Group < ApplicationRecord
has_many :teachers
has_many :students
end

class Teacher < ApplicationRecord
belongs_to :group
has_many :students, :foreign_key => 'group_id', :primary_key => 'group_id'
end

class Student < ApplicationRecord
belongs_to :group
end


And you can retrieve a teacher's students in a group like this:



teacher = Teacher.find(:teacher_id)
students = teacher.students





share|improve this answer





























    0














    Following your example of classes A,B & C and changing only the implementation of class A.



    class A
    belongs_to :b

    def cs
    self.id = b_id
    b_cast = self.becomes(B)
    b_cast.cs
    end
    end


    Console



    A.first                    ==>#<A id: 105, b_id: 1 ...>
    B.first ==>#<B id: 1 ...>
    C.all ==>#<C id: 1, b_id: 1 ...>,
    #<C id: 2, b_id: 1 ...>,
    #<C id: 3, b_id: 1 ...>,
    #<C id: 4, b_id: 1 ...>



    A.first.cs
    A Load (0.2ms) SELECT "as".* FROM "as" ORDER BY "as"."id" ASC LIMIT ? [["LIMIT", 1]]
    C Load (0.1ms) SELECT "cs".* FROM "cs" WHERE "cs"."b_id" = ? LIMIT ? [["b_id", 1], ["LIMIT", 11]]

    A.first.cs ==>#<C id: 1, b_id: 1 ...>,
    #<C id: 2, b_id: 1 ...>,
    #<C id: 3, b_id: 1 ...>,
    #<C id: 4, b_id: 1 ...>


    Official #becomes(klass) documentation can be found
    here!



    On other end, the same effect can be achieved creating an instance of the class B and giving it the id stored in A.first.b_id as follows:



    B.new(id: A.first.b_id).cs
    A Load (0.2ms) SELECT "as".* FROM "as" ORDER BY "as"."id" ASC LIMIT ? [["LIMIT", 1]]
    C Load (0.1ms) SELECT "cs".* FROM "cs" WHERE "cs"."b_id" = ? LIMIT ? [["b_id", 1], ["LIMIT", 11]]

    B.new(id: A.first.b_id).cs ==>#<C id: 1, b_id: 1 ...>,
    #<C id: 2, b_id: 1 ...>,
    #<C id: 3, b_id: 1 ...>,
    #<C id: 4, b_id: 1 ...>





    share|improve this answer





















    • If the solution is necessarily through active records relation point it out so I may remove this answer.
      – MrCodeX
      Nov 21 '18 at 21:37












    • Thank you for the answer. But Solution regarding AR Association is required. Method base approach can be done in multiple ways but it removes all the flexibility which AR associations provide
      – Qaisar Nadeem
      Nov 22 '18 at 20:35











    Your Answer






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    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    1














    An example, Teacher and Student models belong to Group, Teacher has many Student:



    class Group < ApplicationRecord
    has_many :teachers
    has_many :students
    end

    class Teacher < ApplicationRecord
    belongs_to :group
    has_many :students, :foreign_key => 'group_id', :primary_key => 'group_id'
    end

    class Student < ApplicationRecord
    belongs_to :group
    end


    And you can retrieve a teacher's students in a group like this:



    teacher = Teacher.find(:teacher_id)
    students = teacher.students





    share|improve this answer


























      1














      An example, Teacher and Student models belong to Group, Teacher has many Student:



      class Group < ApplicationRecord
      has_many :teachers
      has_many :students
      end

      class Teacher < ApplicationRecord
      belongs_to :group
      has_many :students, :foreign_key => 'group_id', :primary_key => 'group_id'
      end

      class Student < ApplicationRecord
      belongs_to :group
      end


      And you can retrieve a teacher's students in a group like this:



      teacher = Teacher.find(:teacher_id)
      students = teacher.students





      share|improve this answer
























        1












        1








        1






        An example, Teacher and Student models belong to Group, Teacher has many Student:



        class Group < ApplicationRecord
        has_many :teachers
        has_many :students
        end

        class Teacher < ApplicationRecord
        belongs_to :group
        has_many :students, :foreign_key => 'group_id', :primary_key => 'group_id'
        end

        class Student < ApplicationRecord
        belongs_to :group
        end


        And you can retrieve a teacher's students in a group like this:



        teacher = Teacher.find(:teacher_id)
        students = teacher.students





        share|improve this answer












        An example, Teacher and Student models belong to Group, Teacher has many Student:



        class Group < ApplicationRecord
        has_many :teachers
        has_many :students
        end

        class Teacher < ApplicationRecord
        belongs_to :group
        has_many :students, :foreign_key => 'group_id', :primary_key => 'group_id'
        end

        class Student < ApplicationRecord
        belongs_to :group
        end


        And you can retrieve a teacher's students in a group like this:



        teacher = Teacher.find(:teacher_id)
        students = teacher.students






        share|improve this answer












        share|improve this answer



        share|improve this answer










        answered Nov 20 '18 at 7:44









        Gentry Chen

        1,284814




        1,284814

























            0














            Following your example of classes A,B & C and changing only the implementation of class A.



            class A
            belongs_to :b

            def cs
            self.id = b_id
            b_cast = self.becomes(B)
            b_cast.cs
            end
            end


            Console



            A.first                    ==>#<A id: 105, b_id: 1 ...>
            B.first ==>#<B id: 1 ...>
            C.all ==>#<C id: 1, b_id: 1 ...>,
            #<C id: 2, b_id: 1 ...>,
            #<C id: 3, b_id: 1 ...>,
            #<C id: 4, b_id: 1 ...>



            A.first.cs
            A Load (0.2ms) SELECT "as".* FROM "as" ORDER BY "as"."id" ASC LIMIT ? [["LIMIT", 1]]
            C Load (0.1ms) SELECT "cs".* FROM "cs" WHERE "cs"."b_id" = ? LIMIT ? [["b_id", 1], ["LIMIT", 11]]

            A.first.cs ==>#<C id: 1, b_id: 1 ...>,
            #<C id: 2, b_id: 1 ...>,
            #<C id: 3, b_id: 1 ...>,
            #<C id: 4, b_id: 1 ...>


            Official #becomes(klass) documentation can be found
            here!



            On other end, the same effect can be achieved creating an instance of the class B and giving it the id stored in A.first.b_id as follows:



            B.new(id: A.first.b_id).cs
            A Load (0.2ms) SELECT "as".* FROM "as" ORDER BY "as"."id" ASC LIMIT ? [["LIMIT", 1]]
            C Load (0.1ms) SELECT "cs".* FROM "cs" WHERE "cs"."b_id" = ? LIMIT ? [["b_id", 1], ["LIMIT", 11]]

            B.new(id: A.first.b_id).cs ==>#<C id: 1, b_id: 1 ...>,
            #<C id: 2, b_id: 1 ...>,
            #<C id: 3, b_id: 1 ...>,
            #<C id: 4, b_id: 1 ...>





            share|improve this answer





















            • If the solution is necessarily through active records relation point it out so I may remove this answer.
              – MrCodeX
              Nov 21 '18 at 21:37












            • Thank you for the answer. But Solution regarding AR Association is required. Method base approach can be done in multiple ways but it removes all the flexibility which AR associations provide
              – Qaisar Nadeem
              Nov 22 '18 at 20:35
















            0














            Following your example of classes A,B & C and changing only the implementation of class A.



            class A
            belongs_to :b

            def cs
            self.id = b_id
            b_cast = self.becomes(B)
            b_cast.cs
            end
            end


            Console



            A.first                    ==>#<A id: 105, b_id: 1 ...>
            B.first ==>#<B id: 1 ...>
            C.all ==>#<C id: 1, b_id: 1 ...>,
            #<C id: 2, b_id: 1 ...>,
            #<C id: 3, b_id: 1 ...>,
            #<C id: 4, b_id: 1 ...>



            A.first.cs
            A Load (0.2ms) SELECT "as".* FROM "as" ORDER BY "as"."id" ASC LIMIT ? [["LIMIT", 1]]
            C Load (0.1ms) SELECT "cs".* FROM "cs" WHERE "cs"."b_id" = ? LIMIT ? [["b_id", 1], ["LIMIT", 11]]

            A.first.cs ==>#<C id: 1, b_id: 1 ...>,
            #<C id: 2, b_id: 1 ...>,
            #<C id: 3, b_id: 1 ...>,
            #<C id: 4, b_id: 1 ...>


            Official #becomes(klass) documentation can be found
            here!



            On other end, the same effect can be achieved creating an instance of the class B and giving it the id stored in A.first.b_id as follows:



            B.new(id: A.first.b_id).cs
            A Load (0.2ms) SELECT "as".* FROM "as" ORDER BY "as"."id" ASC LIMIT ? [["LIMIT", 1]]
            C Load (0.1ms) SELECT "cs".* FROM "cs" WHERE "cs"."b_id" = ? LIMIT ? [["b_id", 1], ["LIMIT", 11]]

            B.new(id: A.first.b_id).cs ==>#<C id: 1, b_id: 1 ...>,
            #<C id: 2, b_id: 1 ...>,
            #<C id: 3, b_id: 1 ...>,
            #<C id: 4, b_id: 1 ...>





            share|improve this answer





















            • If the solution is necessarily through active records relation point it out so I may remove this answer.
              – MrCodeX
              Nov 21 '18 at 21:37












            • Thank you for the answer. But Solution regarding AR Association is required. Method base approach can be done in multiple ways but it removes all the flexibility which AR associations provide
              – Qaisar Nadeem
              Nov 22 '18 at 20:35














            0












            0








            0






            Following your example of classes A,B & C and changing only the implementation of class A.



            class A
            belongs_to :b

            def cs
            self.id = b_id
            b_cast = self.becomes(B)
            b_cast.cs
            end
            end


            Console



            A.first                    ==>#<A id: 105, b_id: 1 ...>
            B.first ==>#<B id: 1 ...>
            C.all ==>#<C id: 1, b_id: 1 ...>,
            #<C id: 2, b_id: 1 ...>,
            #<C id: 3, b_id: 1 ...>,
            #<C id: 4, b_id: 1 ...>



            A.first.cs
            A Load (0.2ms) SELECT "as".* FROM "as" ORDER BY "as"."id" ASC LIMIT ? [["LIMIT", 1]]
            C Load (0.1ms) SELECT "cs".* FROM "cs" WHERE "cs"."b_id" = ? LIMIT ? [["b_id", 1], ["LIMIT", 11]]

            A.first.cs ==>#<C id: 1, b_id: 1 ...>,
            #<C id: 2, b_id: 1 ...>,
            #<C id: 3, b_id: 1 ...>,
            #<C id: 4, b_id: 1 ...>


            Official #becomes(klass) documentation can be found
            here!



            On other end, the same effect can be achieved creating an instance of the class B and giving it the id stored in A.first.b_id as follows:



            B.new(id: A.first.b_id).cs
            A Load (0.2ms) SELECT "as".* FROM "as" ORDER BY "as"."id" ASC LIMIT ? [["LIMIT", 1]]
            C Load (0.1ms) SELECT "cs".* FROM "cs" WHERE "cs"."b_id" = ? LIMIT ? [["b_id", 1], ["LIMIT", 11]]

            B.new(id: A.first.b_id).cs ==>#<C id: 1, b_id: 1 ...>,
            #<C id: 2, b_id: 1 ...>,
            #<C id: 3, b_id: 1 ...>,
            #<C id: 4, b_id: 1 ...>





            share|improve this answer












            Following your example of classes A,B & C and changing only the implementation of class A.



            class A
            belongs_to :b

            def cs
            self.id = b_id
            b_cast = self.becomes(B)
            b_cast.cs
            end
            end


            Console



            A.first                    ==>#<A id: 105, b_id: 1 ...>
            B.first ==>#<B id: 1 ...>
            C.all ==>#<C id: 1, b_id: 1 ...>,
            #<C id: 2, b_id: 1 ...>,
            #<C id: 3, b_id: 1 ...>,
            #<C id: 4, b_id: 1 ...>



            A.first.cs
            A Load (0.2ms) SELECT "as".* FROM "as" ORDER BY "as"."id" ASC LIMIT ? [["LIMIT", 1]]
            C Load (0.1ms) SELECT "cs".* FROM "cs" WHERE "cs"."b_id" = ? LIMIT ? [["b_id", 1], ["LIMIT", 11]]

            A.first.cs ==>#<C id: 1, b_id: 1 ...>,
            #<C id: 2, b_id: 1 ...>,
            #<C id: 3, b_id: 1 ...>,
            #<C id: 4, b_id: 1 ...>


            Official #becomes(klass) documentation can be found
            here!



            On other end, the same effect can be achieved creating an instance of the class B and giving it the id stored in A.first.b_id as follows:



            B.new(id: A.first.b_id).cs
            A Load (0.2ms) SELECT "as".* FROM "as" ORDER BY "as"."id" ASC LIMIT ? [["LIMIT", 1]]
            C Load (0.1ms) SELECT "cs".* FROM "cs" WHERE "cs"."b_id" = ? LIMIT ? [["b_id", 1], ["LIMIT", 11]]

            B.new(id: A.first.b_id).cs ==>#<C id: 1, b_id: 1 ...>,
            #<C id: 2, b_id: 1 ...>,
            #<C id: 3, b_id: 1 ...>,
            #<C id: 4, b_id: 1 ...>






            share|improve this answer












            share|improve this answer



            share|improve this answer










            answered Nov 21 '18 at 21:23









            MrCodeX

            114




            114












            • If the solution is necessarily through active records relation point it out so I may remove this answer.
              – MrCodeX
              Nov 21 '18 at 21:37












            • Thank you for the answer. But Solution regarding AR Association is required. Method base approach can be done in multiple ways but it removes all the flexibility which AR associations provide
              – Qaisar Nadeem
              Nov 22 '18 at 20:35


















            • If the solution is necessarily through active records relation point it out so I may remove this answer.
              – MrCodeX
              Nov 21 '18 at 21:37












            • Thank you for the answer. But Solution regarding AR Association is required. Method base approach can be done in multiple ways but it removes all the flexibility which AR associations provide
              – Qaisar Nadeem
              Nov 22 '18 at 20:35
















            If the solution is necessarily through active records relation point it out so I may remove this answer.
            – MrCodeX
            Nov 21 '18 at 21:37






            If the solution is necessarily through active records relation point it out so I may remove this answer.
            – MrCodeX
            Nov 21 '18 at 21:37














            Thank you for the answer. But Solution regarding AR Association is required. Method base approach can be done in multiple ways but it removes all the flexibility which AR associations provide
            – Qaisar Nadeem
            Nov 22 '18 at 20:35




            Thank you for the answer. But Solution regarding AR Association is required. Method base approach can be done in multiple ways but it removes all the flexibility which AR associations provide
            – Qaisar Nadeem
            Nov 22 '18 at 20:35


















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