Question

我写了一个课程，并试图打印结果。

当我运行此代码时，我没有收到任何错误，但它没有打印任何内容。它看起来像一片空白。

class Employer

  attr_accessor :id, :first_name, :last_name, :date_joined, :type_of_employment, :vehicle, :location

  def initialize(id, first_name, last_name, date_joined, type_of_employment)
    @employee_id = id
    @employee_first_name = first_name
    @employee_last_name = last_name
    @employee_date_joined = date_joined
    @employee_type_of_employment = type_of_employment
    @vehicle = "Honda"
    @location = "Toronto"
  end

end

emp1obj = Employer.new(1, "John", "Smith", "January 1, 2012", "Fulltime")
puts emp1obj
emp1obj.vehicle = "Honda"
puts emp1obj.first_name
puts emp1obj.last_name
emp1obj.location= "Toronto"

有人可以帮我找到问题吗？

Answer 1

你是如何运行这个文件的？它在这里打印以下输出：

Honda
#<Employer:0x00000001779c68>

请注意，这些是单元化的：

puts emp1obj.first_name
puts emp1obj.last_name

您需要使用：

@first_name = first_name
@last_name = last_name

而不是

@employee_first_name = first_name
@employee_last_name = last_name

在你的构造函数中，如果你想初始化它们。

Answer 2

下面没有输出任何内容，作为 getter 方法的相应实例变量（@first_name，@last_name），:first_name，{ {1}}分别未分配任何值。

:last_name

执行以下操作

puts emp1obj.first_name
puts emp1obj.last_name

根据您的代码@first_name = first_name @last_name = last_name，不需要。

Answer 3

当您开始了解attr_accessor的作用时，这很重要。这就是它正在做的事情：

class Employer

def id
  @employee_id
end

def id=(str)
  @employee_id = str
end

def initialize(id)
    @employee_id = id
end

end
emp1obj = Employer.new(1)
puts emp1obj.id

如果要使用不同的实例变量名，则需要将它们用作方法而不是用于setter / getter的变量。

这对你有用：

class Employer

  attr_accessor :employee_id, :employee_first_name, :employee_last_name, :date_joined, :type_of_employment, :vehicle, :location

def initialize(id, first_name, last_name, date_joined, type_of_employment)
    @employee_id = id
    @employee_first_name = first_name
    @employee_last_name = last_name
    @employee_date_joined = date_joined
    @employee_type_of_employment = type_of_employment
end

end
emp1obj = Employer.new(1, "John", "Smith", "January 1, 2012", "Fulltime")
puts emp1obj.employee_id
puts emp1obj.employee_first_name
puts emp1obj.employee_last_name

类不输出任何结果？

3 个答案: