您的模型创建后FactoryGirl可以生成工厂吗？

Question

当您在dev和Gemfile中的测试块中包含factory_girl_rails gem时,rails会在生成模型时自动生成工厂.

生成模型后有没有办法生成工厂？

---

Eduardo Santana的答案应该是正确的

Answer 1

首先,查看源项目以了解它是如何实现的:

https://github.com/thoughtbot/factory_bot_rails/blob/master/lib/generators/factory_bot/model/model_generator.rb

之后,尝试猜测它是如何工作的:

rails g factory_girl:model Car name speed:integer

结果是:

create  test/factories/cars.rb

内容:

# Read about factories at https://github.com/thoughtbot/factory_girl

FactoryGirl.define do
   factory :car do
     name "MyString"
     speed 1
   end
end

请记住,当您使用rails g时,您可以随时使用rails d撤消它

rails d factory_girl:model Car name speed:integer

Answer 2

该--fixture-replacement选项将让您告诉rails生成用于构建测试数据的内容.您可以将其设置为config/application.rb文件中的默认值,如下所示:

config.generators do |g|
  g.fixture_replacement :factory_girl
end

Answer 3

我有一个宝石正好这个https://github.com/markburns/to_factory

Answer 4

这对我来说很有效，使用rails g factory_bot:model 用户可以运行命令，也可以只是将命令输出。您仍然需要填写该值。

@run_command        = true
@force              = true
@columns_to_ignore  = %w[id created_at update_at]
@tables_to_ignore = %w[schema_migrations ar_internal_metadata]
tables = ActiveRecord::Base.connection.tables.reject{|t| (@tables_to_ignore || []).include?(t)}

tables.each do |table|
  klass = table.singularize.camelcase.constantize
    command = "rails g factory_bot:model #{klass.to_s} #{klass.columns.reject do |c|
      (@columns_to_ignore || []).include?(c.name)
    end.map do |d|
      "#{d.name}:#{d.sql_type == 'jsonb' ? 'json' : d.type}"
  end.join(' ')}"
  command << ' --force' if @force
  puts command
  puts %x{#{command}} if @run_command
  puts (1..200).to_a.map{}.join('-')
end