Question

我有一个membership模型，它具有以下关联：

class Membership < ActiveRecord::Base
  belongs_to :family_tree
  belongs_to :member
  belongs_to :inviter, class_name: "User", foreign_key: "user_id"
  belongs_to :invited, class_name: "User", foreign_key: "invited_id"    
end

这是架构：

# == Schema Information
#
# Table name: memberships
#
#  id                      :integer          not null, primary key
#  family_tree_id          :integer
#  user_id                 :integer
#  relation                :string(255)
#  member_id               :integer
#  invited_id              :integer
#

在我的User型号上，我有以下内容：

class User < ActiveRecord::Base
 has_many :memberships
end

所以我可以成功地将2个不同的用户分配给1个记录，例如：

[7] pry(#<RSpec::ExampleGroups::User>)> membership
=> #<Membership id: 90, family_tree_id: 1049, user_id: 779, created_at: "2015-10-29 15:24:02", updated_at: "2015-10-29 15:24:02", relation: "sister", member_id: 108, connection_sent_at: nil, connection_responded_at: nil, connect_send_limit: nil, connect_times_sent: nil, connected: nil, connect_type: nil, request_status: nil, connection_removed_at: nil, invited_id: 777>
[8] pry(#<RSpec::ExampleGroups::User>)> membership.family_tree
=> #<FamilyTree id: nil, name: "Enrico McGlynn", user_id: 777, created_at: "2015-10-29 15:24:02", updated_at: "2015-10-29 15:24:02">
[11] pry(#<RSpec::ExampleGroups::User>)> membership.inviter
=> #<User id: nil, email: "odell@lowe.net", encrypted_password: "$2a$04$Ke.1sQ9mZ2ZP0Hxamg6NS.6zUoEHyolPbKJ1UfAlils...", reset_password_token: nil, reset_password_sent_at: nil, remember_created_at: nil, sign_in_count: 0, current_sign_in_at: nil, last_sign_in_at: nil, current_sign_in_ip: nil, last_sign_in_ip: nil, created_at: "2015-10-29 15:24:02", updated_at: "2015-10-29 15:24:02", first_name: "Audie", confirmation_token: nil, confirmed_at: "2015-10-29 15:23:57", confirmation_sent_at: nil, unconfirmed_email: nil, invitation_relation: "quibusdam", avatar: nil, invitation_token: nil, invitation_created_at: nil, invitation_sent_at: nil, invitation_accepted_at: nil, invitation_limit: nil, invited_by_id: nil, invited_by_type: nil, invitations_count: 0, bio: "Quibusdam est laborum cupiditate modi et est. Nece...", last_name: "Feil", gender: 1>
[12] pry(#<RSpec::ExampleGroups::User>)> membership.invited
=> #<User id: nil, email: "jalon_greenfelder@hudson.name", encrypted_password: "$2a$04$jlFrLS6UlcchaJlsOAv4ne/5nYSxH.7Zdsvk5WZMd1w...", reset_password_token: nil, reset_password_sent_at: nil, remember_created_at: nil, sign_in_count: 0, current_sign_in_at: nil, last_sign_in_at: nil, current_sign_in_ip: nil, last_sign_in_ip: nil, created_at: "2015-10-29 15:24:02", updated_at: "2015-10-29 15:24:02", first_name: "Annamae", confirmation_token: nil, confirmed_at: "2015-10-29 15:23:57", confirmation_sent_at: nil, unconfirmed_email: nil, invitation_relation: "eveniet", avatar: nil, invitation_token: nil, invitation_created_at: nil, invitation_sent_at: nil, invitation_accepted_at: nil, invitation_limit: nil, invited_by_id: nil, invited_by_type: nil, invitations_count: 0, bio: "Ut explicabo omnis facilis et. Dolor sunt enim et....", last_name: "Kshlerin", gender: 1>

我遇到的问题是当我试图走另一条路时，即从该用户记录中找到与之关联的所有memberships。它返回一个空数组。

[1] pry(#<RSpec::ExampleGroups::User>)> u1
=> #<User id: nil, email: "jalon_greenfelder@hudson.name", encrypted_password: "$2a$04$jlFrLS6UlcchaJlsOAv4ne/5nYSxH.7Zdsvk5WZMd1w...", reset_password_token: nil, reset_password_sent_at: nil, remember_created_at: nil, sign_in_count: 0, current_sign_in_at: nil, last_sign_in_at: nil, current_sign_in_ip: nil, last_sign_in_ip: nil, created_at: "2015-10-29 15:24:02", updated_at: "2015-10-29 15:24:02", first_name: "Annamae", confirmation_token: nil, confirmed_at: "2015-10-29 15:23:57", confirmation_sent_at: nil, unconfirmed_email: nil, invitation_relation: "eveniet", avatar: nil, invitation_token: nil, invitation_created_at: nil, invitation_sent_at: nil, invitation_accepted_at: nil, invitation_limit: nil, invited_by_id: nil, invited_by_type: nil, invitations_count: 0, bio: "Ut explicabo omnis facilis et. Dolor sunt enim et....", last_name: "Kshlerin", gender: 1>
[9] pry(#<RSpec::ExampleGroups::User>)> u1.memberships
=> []

如何让user.memberships在user_id and invited_id模型上返回与memberships相关联的成员资格？

Answer 1

Membership.where(user_id: user.id)相当于memberships。您应该注意User中与user_id模型关联的foreign_key默认为class User < ActiveRecord::Base #user.memberships will be equivalent to Membership.where(user_id: user.id) has_many :memberships #user.invited_memberships will be equivalent to Membership.where(inviter_id: user.id) has_many :invited_memberships, class_name: Membership, foreign_key: :inviter_id def all_memberships #this will return the memberships associated with the user_id and the inviter_id #you may want to move this to the Membership model Membership.where('user_id = ? OR inviter_id = ?', self.id, self.id) end end。要解决您的问题：

m <- regexpr(paste0(x,collapse=""),paste0(z,collapse = ""),fixed=T)
seq(m,length.out=length(x))

如何在这个foreign_key协会上获得关联的反面？

1 个答案: