Question

我有一个帮助方法，可以为资源生成“crud”链接。现在它只处理编辑和销毁。但我想重构它，以便在传递模型类而不是实例时创建new链接。

期望的输出：

> crud_links_for_resource(Product)
> { :new => '<a href="/products/new">Create a new product.</a>'}

检查变量是模型类还是实例的最佳方法是什么？我想过使用duck typing（resource.respond_to? :new_record?）但是有更好的方法吗？

module PermissionsHelper
  # Generates a hash of links to edit or destroy a resource
  # @param [ActiveModel] resource
  # @param [Actions] a list of crud actions to create links to
  # @param [Hash] kwargs optional hash to pass to link to
  # @option kwargs [String] :controller - controller name to use.
  #   Otherwise a guess is performed based on the resource class name.
  # @option kwargs [Hash] url_extras - passed to url_for. Can be used for nested resources.
  # @return [Hash] a list of links to actions which the user is allowed to perform
  def crud_links_for_resource(resource,  actions = [:destroy, :edit], **kwargs)
    privledges = actions.keep_if { |action| can? action, resource }
    privledges.each_with_object({}) do |action, hash|
      i18n_key = resource.model_name.i18n_key
      txt = t("#{ i18n_key }.#{action}")
      controller = kwargs[:controller] || i18n_key.to_s.pluralize
      url_extras = kwargs[:url_extras] || {}
      options = kwargs.except(:controller, :url_extras)
      case action
        when :destroy
          options.merge!(method: :delete, confirm: t("#{ i18n_key }.confirm_#{action}"))
        else
      end
      hash[action] = link_to(txt, { action: action, controller: controller, id: resource }.merge(url_extras) ,options)
    end
  end
end

Answer 1

if myobject.is_a?(Class)
  #it's a class
else
  #it's not
end

或

create

Answer 2

如果你想确保有问题的对象不仅是任何类而是模型类 - 即ActiveRecord::Base的<直接或间接的]后代：

my_object.is_a?(Class) && my_object < ActiveRecord::Base

如何检查参数是模型的实例还是Rails中的类？

2 个答案: