检查控制器是否在Rails中有一个模型-CheckifcontrollerhasamodelinRails

2  

You could try something like this, if you pass in the name of the controller as a string. This solution assumes that your models are using ActiveRecord prior to rails 5 where ApplicationRecord was used to define models; in that case just switch ActiveRecord::Base with ApplicationRecord. Also if you have models that are plain old ruby objects (POROs), then this wont work for them.

如果您将控制器的名称作为字符串传入，您可以尝试类似的操作。这个解决方案假设您的模型使用的是ActiveRecord，而rails 5使用ApplicationRecord来定义模型;在这种情况下，只需用ApplicationRecord交换ActiveRecord: Base。此外，如果您拥有的模型是普通的旧ruby对象(POROs)，那么这对它们就不起作用了。

def has_model?(controller_klass)
  begin
    class_string = controller_klass.to_s.gsub('Controller', '').singularize
    class_instance = class_string.constantize.new
    return class_instance.class.ancestors.include? ActiveRecord::Base 
  rescue NameError => e
    return false
  end
end 

0  

This method doesn't rely on exceptions, and works with input as Class or String. It should work for any Rails version :

该方法不依赖于异常，并将输入作为类或字符串处理。它应该适用于任何Rails版本:

def has_model?(controller_klass)
  all_models = ActiveRecord::Base.descendants.map(&:to_s)
  model_klass_string = controller_klass.to_s.sub(/Controller$/,'').singularize
  all_models.include?(model_klass_string)
end

Note : you need to set

注意:您需要设置。

config.eager_load = true

in

在

config/environments/development.rb

配置/环境/ development.rb

If you have non ActiveRecord models, you can ignore the previous note and use :

如果您有非ActiveRecord模型，您可以忽略之前的注释并使用:

all_models = Dir[File.join(Rails.root,"app/models", "**","*.rb")].map{|f| File.basename(f,'.rb').camelize}