关系模型未按预期工作
关系模型未按预期工作
提问于 2018-06-04 05:38:08
我使用迁移创建了以下Active Record架构，但关系与架构不对应。我尝试过重置、删除、创建和迁移，但在Rails C中，如果我创建了一个用户方法！(...)，然后查询u.groups或u.genres，我得到了'undefined u.User.create‘
谢谢你的帮忙
ActiveRecord::Schema.define(version: 20180603211047) do
# These are extensions that must be enabled in order to support this database
enable_extension "plpgsql"
create_table "genres", force: :cascade do |t|
t.datetime "created_at", null: false
t.datetime "updated_at", null: false
t.string "tag"
t.bigint "user_id"
t.index ["user_id"], name: "index_genres_on_user_id"
end
create_table "genres_users", id: false, force: :cascade do |t|
t.bigint "user_id", null: false
t.bigint "genre_id", null: false
end
create_table "groups", force: :cascade do |t|
t.datetime "created_at", null: false
t.datetime "updated_at", null: false
t.string "name"
t.bigint "user_id"
t.index ["user_id"], name: "index_groups_on_user_id"
end
create_table "groups_users", id: false, force: :cascade do |t|
t.bigint "user_id", null: false
t.bigint "group_id", null: false
end
create_table "playlists", force: :cascade do |t|
t.datetime "created_at", null: false
t.datetime "updated_at", null: false
t.string "name"
t.string "link"
t.text "description"
t.bigint "group_id"
t.index ["group_id"], name: "index_playlists_on_group_id"
end
create_table "users", force: :cascade do |t|
t.datetime "created_at", null: false
t.datetime "updated_at", null: false
t.string "email", default: "", null: false
t.string "encrypted_password", default: "", null: false
t.string "reset_password_token"
t.datetime "reset_password_sent_at"
t.datetime "remember_created_at"
t.integer "sign_in_count", default: 0, null: false
t.datetime "current_sign_in_at"
t.datetime "last_sign_in_at"
t.string "current_sign_in_ip"
t.string "last_sign_in_ip"
t.string "name"
t.string "token"
t.date "birthday"
t.string "link"
t.string "playlistId"
t.string "country"
t.index ["email"], name: "index_users_on_email", unique: true
t.index ["reset_password_token"], name: "index_users_on_reset_password_token", unique: true
end
add_foreign_key "genres", "users"
add_foreign_key "groups", "users"
add_foreign_key "playlists", "groups"
end以下是模型：
class User < ApplicationRecord
# Include default devise modules. Others available are:
# :confirmable, :lockable, :timeoutable and :omniauthable
devise :database_authenticatable, :registerable,
:recoverable, :rememberable, :trackable, :validatable
#before_action :authenticate_user!
has_and_belongs_to_many :genres, :through => :genres_users
has_and_belongs_to_many :groups, :through => :groups_users
include Enumerable
end
class Genre < ApplicationRecord
has_and_belongs_to_many :users, :through => :genres_users
end
class Group < ApplicationRecord
has_and_belongs_to_many :users, :through => :groups_users
has_one :playlist
end
class Playlist < ApplicationRecord
belongs_to :group
end这种关系是组有用户，用户有流派(最喜欢的流派！)，这些都是通过连接表(每个用户有多个流派，每个用户有多个组)具有并属于关系。每个组都有一个播放列表，并且会有多个播放列表
回答 1
Stack Overflow用户
发布于 2018-06-04 06:01:05
从操作中澄清后编辑
关系是组有用户，用户有流派(最喜欢的流派！)，这些都是通过连接表(每个用户有多个流派，每个用户有多个组)的关系。每个组都有一个播放列表，并且会有多个播放列表
首先，您不需要关于组或流派的user_id专栏，因为这不应该是设置的工作方式。
class Genre < ApplicationRecord
has_many :favorite_genres
has_many :users, through: :favorite_genres
[... other stuff]
end
class User < ApplicationRecord
has_many :group_memberships
has_many :groups, through: :group_memberships
has_many :favorite_genres
has_many :users, through: :favorite_genres
[... other stuff]
end
class Group < ApplicationRecord
has_many :group_memberships
has_many :users, through: :group_memberships
has_many :playlists
[... other stuff]
end
class Playlist < ApplicationRecord
belongs_to :group
end
class GroupMemberships < ApplicationRecord
belongs_to :user
belongs_to :group
[... other stuff]
end
class FavoriteGenres < ApplicationRecord
belongs_to :user
belongs_to :genre
[... other stuff]
end因此，您将以组的形式删除user_id列。连接发生在:group_memberships (表以前称为users_groups)中，它是一个user_id、一个group_id，然后您可以根据需要拥有额外的元数据列(例如admin boolean/group_id等)。这就是所谓的“多点直通”关系(http://guides.rubyonrails.org/association_basics.html#the-has-many-through-association)。
同样，用户最喜欢的流派也是通过关系设置的。因此，您将有一个单独的数据库表和模型文件用于这些通过连接。
我认为在这个级别上根本不需要add_foreign_key调用，也不需要很多索引。您可能会执行更多的急切加载，或者可能在thorugh join表上添加索引，并且您将在模式中执行以下操作：
t.index ["user_id", "genre_id"], name: "index_favorite_genres_on_user_id_and_genre_id"请记住，belongs_to现在为它创建了一个在5.x中存在的验证。您可以通过在模型中的该行上添加optional: true来覆盖它，例如belongs_to :foo, optional: true
所以，这就是你的新模式：
create_table "genres", id: :serial, force: :cascade do |t|
t.string "tag"
t.datetime "created_at", null: false
t.datetime "updated_at", null: false
end
create_table "groups", id: :serial, force: :cascade do |t|
t.string "name"
t.datetime "created_at", null: false
t.datetime "updated_at", null: false
end
create_table "favorite_genres", id: false, force: :cascade do |t|
t.bigint "user_id", null: false
t.bigint "genre_id", null: false
end
create_table "groups_memberships", id: false, force: :cascade do |t|
t.bigint "user_id", null: false
t.bigint "group_id", null: false
end
create_table "playlists", id: :serial, force: :cascade do |t|
t.datetime "created_at", null: false
t.datetime "updated_at", null: false
t.string "name"
t.string "link"
t.text "description"
t.bigint "group_id"
t.index ["group_id"], name: "index_playlists_on_group_id"
end
create_table "users", id: :serial, force: :cascade do |t|
t.datetime "created_at", null: false
t.datetime "updated_at", null: false
t.string "email", default: "", null: false
t.string "encrypted_password", default: "", null: false
t.string "reset_password_token"
t.datetime "reset_password_sent_at"
t.datetime "remember_created_at"
t.integer "sign_in_count", default: 0, null: false
t.datetime "current_sign_in_at"
t.datetime "last_sign_in_at"
t.string "current_sign_in_ip"
t.string "last_sign_in_ip"
t.string "name"
t.string "token"
t.date "birthday"
t.string "link"
t.string "playlistId"
t.string "country"
t.index ["email"], name: "index_users_on_email", unique: true
t.index ["reset_password_token"], name: "index_users_on_reset_password_token", unique: true
end试一试(我还没有在应用程序中构建它，所以代码中可能有一些错误)，你现在应该能够运行你的控制台了：
u = User.create([values])
u.genres (should return nil until you create some relationships)等。
页面原文内容由Stack Overflow提供。腾讯云小微IT领域专用引擎提供翻译支持
原文链接：
https://stackoverflow.com/questions/50671212
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