ruby-on-rails - Rails 如何仅从 ActiveRecord 对象中提取(和复制)某些属性？

Question

在 Rails 5 中，我想创建一个 Author 模型的新实例，但在此过程中，我只想从我的 User 模型中复制属性值(列)的一个子集。

我正在尝试运行以下代码:

user = User.first
Author.create user.attributes.without("id", "token", "secret", "created_at", "updated_at", "active", "email", "encrypted_password", "first_name", "last_name", "reset_password_token", "reset_password_sent_at", "remember_created_at", "sign_in_count", "current_sign_in_at", "last_sign_in_at", "current_sign_in_ip", "last_sign_in_ip")

User.first 方法有效并从表中返回一个有效用户。但是， Author.create 行不起作用。我收到一个 NoMethodError: undefined method `each' for nil:NilClass 错误。

这是我来自 schema.rb 文件的用户表:

create_table "users", force: :cascade do |t|
    t.string   "provider",                               null: false
    t.string   "uid",                                    null: false
    t.string   "name"
    t.string   "location"
    t.string   "image_url"
    t.string   "url"
    t.string   "token"
    t.string   "secret"
    t.datetime "created_at",                             null: false
    t.datetime "updated_at",                             null: false
    t.boolean  "active",                 default: false
    t.text     "description"
    t.string   "screen_name"
    t.integer  "followers"
    t.integer  "following"
    t.integer  "statuses"
    t.integer  "listed"
    t.integer  "lists"
    t.integer  "favorites"
    t.integer  "views",                  default: 0
    t.string   "website"
    t.string   "email",                  default: "",    null: false
    t.string   "encrypted_password",     default: "",    null: false
    t.string   "first_name",             default: "",    null: false
    t.string   "last_name",              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.inet     "current_sign_in_ip"
    t.inet     "last_sign_in_ip"
    t.index ["active"], name: "index_users_on_active", using: :btree
    t.index ["email"], name: "index_users_on_email", unique: true, using: :btree
    t.index ["provider", "uid"], name: "index_users_on_provider_and_uid", unique: true, using: :btree
    t.index ["provider"], name: "index_users_on_provider", using: :btree
    t.index ["uid"], name: "index_users_on_uid", using: :btree
end

这是我的作者表:

create_table "authors", force: :cascade do |t|
    t.string   "uid",                     null: false
    t.string   "provider",                null: false
    t.string   "name"
    t.string   "location"
    t.string   "image_url"
    t.string   "url"
    t.text     "description"
    t.string   "screen_name"
    t.integer  "followers",   default: 0
    t.integer  "following",   default: 0
    t.integer  "statuses",    default: 0
    t.integer  "listed",      default: 0
    t.integer  "lists",       default: 0
    t.integer  "favorites",   default: 0
    t.integer  "views",       default: 0
    t.string   "website"
    t.datetime "created_at",              null: false
    t.datetime "updated_at",              null: false
    t.index ["provider"], name: "index_authors_on_provider", using: :btree
    t.index ["uid", "provider"], name: "index_authors_on_uid_and_provider", unique: true, using: :btree
    t.index ["uid"], name: "index_authors_on_uid", using: :btree
end

如您所见，新的 Author 表具有许多与 User 表相同的列/属性，但我只想在创建新的 Author 对象时复制 User 属性(列和值)的子集。

也许 user.attributes.without 方法不是正确的方法？

Answer 1

您可以通过 slice 来完成自 user.attributes只是值的散列。

user.attributes.slice('name', 'location')