Question

我有用户表：

class User < ActiveRecord::Base
  has_many :broker_clients, :class_name => "BrokerClients", :foreign_key => "broker_id"
  has_many :clients, :through => :broker_clients, :foreign_key => "broker_id"
  has_many :brokers, :through => :broker_clients, :foreign_key => "client_id"
end

和BrokerClients表：

class BrokerClients < ActiveRecord::Base
  belongs_to :broker, class_name: "User"
  belongs_to :client, class_name: "User"
end

现在我创建了一段关系：

>> BrokerClients.create(broker_id: User.first.id, client_id: User.last.id)
  User Load (9.7ms)  SELECT "users".* FROM "users" ORDER BY "users"."id" ASC LIMIT 1
  User Load (1.3ms)  SELECT "users".* FROM "users" ORDER BY "users"."id" DESC LIMIT 1
   (1.3ms)  BEGIN
  SQL (41.5ms)  INSERT INTO "broker_clients" ("broker_id", "client_id", "created_at", "updated_at") VALUES ($1, $2, $3, $4) RETURNING "id"  [["broker_id", 4], ["client_id", 210], ["created_at", Fri, 10 Oct 2014 13:43:27 EDT -04:00], ["updated_at", Fri, 10 Oct 2014 13:43:27 EDT -04:00]]
   (0.5ms)  COMMIT
=> #<BrokerClients id: 1, broker_id: 4, client_id: 210, created_at: "2014-10-10 17:43:27", updated_at: "2014-10-10 17:43:27">
>> User.first.brokers.first

当我试图让客户正常工作时：

>> User.first.clients.first
  User Load (0.6ms)  SELECT "users".* FROM "users" ORDER BY "users"."id" ASC LIMIT 1
  User Load (1.2ms)  SELECT "users".* FROM "users" INNER JOIN "broker_clients" ON "users"."id" = "broker_clients"."
=> #<User id: 210, ....

但对于客户来说，当我试图让与之相关的经纪人无法工作时：

>> User.last.brokers.first
  User Load (0.7ms)  SELECT "users".* FROM "users" ORDER BY "users"."id" DESC LIMIT 1
  User Load (0.7ms)  SELECT "users".* FROM "users" INNER JOIN "broker_clients" ON "users"."id" = "broker_clients"."broker_id" WHERE "broker_clients"."broker_id" = $1 ORDER BY "users"."id" ASC LIMIT 1  [["broker_id", 210]]
=> nil

有任何帮助吗？

Answer 1

将User modal更改为： -

has_many :broker_clients, :class_name => "BrokerClients", :foreign_key => "broker_id"
has_many :clients, :through => :broker_clients, :foreign_key => "broker_id"

has_many :inverse_broker_clients, :class_name => "BrokerClients", :foreign_key => "client_id"
has_many :brokers, :through => :inverse_broker_clients, :foreign_key => "client_id"

Answer 2

查看您的插入内容与您的查询：

插入：

SQL（41.5ms）INSERT INTO“broker_clients”（“broker_id”，“client_id”，“created_at”，“updated_at”）VALUES（$ 1，$ 2，$ 3，$ 4）返回“id”[[“broker_id”， 4]，[“client_id”，210]，[“created_at”，星期五，2014年10月10日13:43:27 EDT -04：00]，[“updated_at”，星期五，2014年10月10日13:43:27美国东部时间 - 04:00]

查询：

用户负载（0.7ms）SELECT“users”。* FROM“users”INNER JOIN“broker_clients”ON“users”。“id”=“broker_clients”。“broker_id”WHERE“broker_clients”。“broker_id”= $ 1 ORDER BY“users”。“id”ASC LIMIT 1 [[“broker_id”，210]]

您插入（broker_id = 4，client_id = 210）但查询broker_id = 210。这就是你为brokers.first获得nil的原因。很确定你的外键是倒退的：

has_many :clients, :through => :broker_clients, :foreign_key => "broker_id"
has_many :brokers, :through => :broker_clients, :foreign_key => "client_id"

应该是

has_many :clients, :through => :broker_clients, :foreign_key => "client_id"
has_many :brokers, :through => :broker_clients, :foreign_key => "broker_id"

希望有所帮助。

has_many通过与同一模型的关系

2 个答案: