I have the following two models:
class Message < ActiveRecord::Base
belongs_to :to_user, :class_name => 'User'
belongs_to :from_user, :class_name => 'User'
has_ancestry #Using the 'ancestry' gem
end
class User < ActiveRecord::Base
has_many :messages_received, :class_name => 'Message', :foreign_key => 'to_user_id'
has_many :messages_sent, :class_name => 'Message', :foreign_key => 'from_user_id'
end
Each user is allowed to have one conversation with another user and all the replies should be threaded from the original message.
In my 'index' controller action how do I query both sent messages and received messages? For example, if User1 hits '/users/2/messages/' they should see the whole conversation between user1 and user2 (regardless of who sent the first message). Do I need to add a 'Thread' model or is there a way to accomplish this with my current structure?
Thanks.
You may be better off restructuring this as a conversation to which you can join people than a series of interconnected messages in a chain. For instance:
class Conversation < ActiveRecord::Base
has_many :messages
has_many :participants
has_many :users, :through => :participants
end
class Message < ActiveRecord::Base
belongs_to :conversation
end
class Participant < ActiveRecord::Base
belongs_to :conversation
belongs_to :user
end
class User < ActiveRecord::Base
has_many :conversations
has_many :participants
end
When a message is sent to someone, create a conversation for it and invite the appropriate parties by adding them to the users
list.
Threaded messaging could be added here by building a parent relationship into the Message itself or using ancestry, though in practice this tends to be over-kill as simple chronological ordering of replies is usually sufficient for most people.
To track read/unread status you will need an association table between user and messages directly, and this can be tricky, so avoid it unless you need it.
Keep in mind that some names are reserved by either Ruby or Rails, and Thread
is one of them, so you can't have a model with that name.
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