在一个模型上调用一个块

Question

不像这样在类中定义作用域:

 scope :first_user, -> { first }

这样命名:User.first_user

我想在另一个类中定义一个块，它可以在用户类中调用，并像Scope一样工作:

这段代码不工作，但它应该表明我想要实现的行为:

class Manage
 def get_first_user
    User.&first_added
 end
 def first_added
   Proc.new { first }
 end
end 

当我运行这段代码时:

a = Manage.new
a.get_first_user

上面写着我，& undefined method for User。如何在User模型上执行first_added中定义的Block ?

通常如何在类上调用块?谢谢

Answer 1

如果我正确理解了您的问题，您可以使用class_eval:

>> foo = Proc.new { count }
=> #<Proc:0x007f1aa7cacfd8@(pry):30>
>> Buyer.class_eval(&foo)
(30.6ms)  SELECT COUNT(*) FROM "buyers"
=> 1234

或者用你的例子:

class Manage
  def get_first_user
    User.class_eval(&first_added)
  end
  def first_added
    Proc.new { first }
  end
end 

Answer 2

这不是你想要的，但可能会有帮助。
我不确定是否可以在上调用进程。我认为一个人只能用来调用proc ，比如在你的情况下传递User作为参数。

def get_first_user
  wrap User, &first_added
end
def first_added
  Proc.new { |model| model.where(...) }
end
private
def wrap(model, &block)
  yield model
end

Answer 3

从Manager调用User上的first有三种方法:

• Object#send:http://ruby-doc.org/core-2.3.1/Object.html method-i-send

• Module#class_eval:http://ruby-doc.org/core-2.3.1/Module.html method-i-class_eval

File manage.rb:

class User
  def self.first
    puts 'record would probably go here'
  end
  def self.yielder
    print "via yielder => "
    self.send(yield) if block_given?
  end
  def self.blocker(&block)
    print "via blocker => "
    self.send(block.call) if block_given?
  end
  def self.evaller(&block)
    print "via evaller => "
    self.class_eval(block.call) if block_given?
  end
end
class Manage
  def get_first_user
    User.yielder(&first_added)
    User.blocker(&first_added)
    User.evaller(&first_added)
  end
  def first_added
    Proc.new {"first"} 
  end 
end
a = Manage.new
a.get_first_user

输出:

$ ruby manage.rb 
via yielder => record would probably go here
via blocker => record would probably go here
via evaller => record would probably go here