我正在尝试创建一个与has_one / belongs_to相关的FactoryBot对象
用户has_one Car
Car has_one Style
Style有一个属性{style_number:“1234”}
我的控制器引用用户,用户has_one Car,Car has_one Style,我需要在FactoryBot中设置这些值。
如何创建具有Car对象的User,该User具有Style对象?
我阅读了文档https://github.com/thoughtbot/factory_bot/blob/master/GETTING_STARTED.md
但是,我不明白他们如何推荐这样做,我想我需要嵌套三个对象,但对语法感到困惑。
before_action :authenticate_user!
before_action :set_steps
before_action :setup_wizard
include Wicked::Wizard
def show
@user = current_user
@form_object = form_object_model_for_step(step).new(@user)
render_wizard
end
私人的
def set_steps
if style_is_1234
self.steps = car_steps.insert(1, :style_car)
else
self.steps = car_steps
end
end
def style_is_1234
if params.dig(:form_object, :style_number)
(params.dig(:form_object, :style_number) & ["1234"]).present?
else
(current_user.try(:car).try(:style).try(:style_number) & ["1234"]).present?
end
end
def car_steps
[:type,:wheel, :brand]
end
FactoryBot.define do
factory :user, class: User do
first_name { "John" }
last_name { "Doe" }
email { Faker::Internet.email }
password { "somepassword" }
password_confirmation { "some password"}
end
end
before(:each) do
@request.env["devise.mapping"] = Devise.mappings[:user]
user = FactoryBot.create(:user)
sign_in user
context "Requesting with second step CarStyle" do
it "should return success" do
get :show, params: { :id => 'car_style' }
expect(response.status).to eq 200
end
end
目前此测试失败,因为User.Car.Style.style_number未设置为“1234”。
FactoryBot.define do
factory :user, class: User do
first_name { "John" }
last_name { "Doe" }
email { Faker::Internet.email }
password { "somepassword" }
password_confirmation { "some password"}
car
end
end
FactoryBot.define do
factory :car, class: Car do
make { "Holden" }
model { "UTE" }
end
end
FactoryBot.define do
factory :style, class: Style do
color { "blue" }
for_car
trait :for_car do
association(:styable, factory: :car)
end
end
end
SystemStackError:堆栈级别太深
我试过srng的推荐
编辑:对于多态关联尝试;
FactoryBot.define做工厂:汽车,类:汽车做{“Holden”}模型{“UTE”}协会:stylable,工厂:: style end end
并得到错误:
ActiveRecord :: RecordInvalid:验证失败:样式必须存在
我认为这是一个轨道5问题。 https://github.com/rails/rails/issues/24518
但是,我想保留我的代码添加可选:true。有什么办法吗?
FactoryBot.define do
factory :car, class: Car do
make { "Holden" }
model { "UTE" }
after(:create) do |car|
create(:style, stylable: car)
end
end
end
试过Srng的第二个推荐,虽然它对他有用,但我得到了一个稍微不同的错误:
ActiveRecord :: RecordInvalid:验证失败:用户必须存在
为了创建依赖的工厂,您必须为每个模型创建一个工厂,然后只需将依赖的Model
名称添加到您的工厂,即。
FactoryBot.define do
factory :user, class: User do
first_name { "John" }
last_name { "Doe" }
email { Faker::Internet.email }
password { "somepassword" }
password_confirmation { "some password"}
car
end
end
FactoryBot.define do
factory :car, class: Car do
make { "Holden" }
model { "UTE" }
style
end
end
FactoryBot.define do
factory :style, class: Style do
color { "blue" }
end
end
编辑:相关代码;
# Factories
FactoryBot.define do
factory :user, class: User do
first_name { "John" }
last_name { "Doe" }
email { Faker::Internet.email }
password { "somepassword" }
password_confirmation { "some password"}
after(:create) do |user|
user.car ||= create(:car, :user => user)
end
end
end
factory :style, class: Style do
style_number { "Blue" }
end
factory :car, class: Car do
name { "Holden" }
trait :style do
association :stylable, factory: :style
end
end
#models
class Car < ApplicationRecord
has_one :style, as: :styleable
end
class Style < ApplicationRecord
belongs_to :styleable, polymorphic: true
belongs_to :car
end
# Migrations - The belongs_to is the only important one
class CreateStyles < ActiveRecord::Migration[5.2]
def change
create_table :styles do |t|
t.string :style_number
t.belongs_to :stylable, polymorphic: true
t.timestamps
end
end
end
class CreateCars < ActiveRecord::Migration[5.2]
def change
create_table :cars do |t|
t.string :name
t.timestamps
end
end
end