SQL查询通过相关模型

Question

我的问题是选择一个用户的朋友？

红宝石如下：

class User < ActiveRecord::Base 
    has_many :relations 
end 

class Relation < ActiveRecord::Base 
    belongs_to :initiator, class_name: "User" 
    belongs_to :initiatee, class_name: "User" 
end 

class CreateRelations < ActiveRecord::Migration 
    def change 
    create_table :relations do |t| 
     t.references :initiator 
     t.references :initiatee 
     t.boolean  :is_confirmed, default: false 
     t.integer  :type 
    end 
    end 
end 

Answer 1

你会遇到麻烦，因为Rails的预计type用于Single Table Inheritance。您还需要告诉Rails，relations上的ID不是user_id，这将是has_many的默认值。既然你有两个关系方向，你就需要声明两者。

has_many :outgoing_relations, class_name: 'Relation', foreign_key: 'initiator_id' 
    has_many :incoming_relations, class_name: 'Relation', foreign_key: 'initiatee_id' 

从那里，做最简单的事情就是写一个聚合了其他用户的方法：

def friends(params = {}) 
    outgoing_relations.where(params).includes(:initiatee).map(&:initiatee) + 
    incoming_relations.where(params).includes(:initiator).map(&:initiator) 
end 

> User.first.friends(is_confirmed: true, kind: 0) 
=> [#<User id: 2, created_at: "2015-08-28 15:11:12", updated_at: "2015-08-28 15:11:12">] 

在直SQL，你可以很容易地UNION一对夫妇查询拉您想要的其他用户ID，然后按照您喜欢的方式进行操作。

SELECT initiatee_id AS id 
FROM relations 
WHERE initiator_id = 2 
    AND kind = 0 
    AND is_confirmed 
UNION 
SELECT initiator_id AS id 
FROM relations 
WHERE initiatee_id = 2 
    AND kind = 0 
    AND is_confirmed 
; 

id 
---- 
    1 
    3 

这是数据我上运行：

SELECT * FROM users; 
id |   created_at   |   updated_at 
----+----------------------------+---------------------------- 
    1 | 2015-08-28 15:11:10.631187 | 2015-08-28 15:11:10.631187 
    2 | 2015-08-28 15:11:12.911575 | 2015-08-28 15:11:12.911575 
    3 | 2015-08-28 15:14:27.762946 | 2015-08-28 15:14:27.762946 

SELECT * FROM relations; 
id | initiator_id | initiatee_id | is_confirmed | kind 
----+--------------+--------------+--------------+------ 
    1 |   1 |   2 | t   | 0 
    2 |   3 |   2 | t   | 0