Question

用详细的例子重写问题：

我有三个属性的对象数组，我想构建苛刻的数组，并且得到的哈希没有@name实例属性。我用一个例子模拟了这个问题。

示例创建

class Employee
  attr_accessor :name, :company, :duration
  def initialize(name, company, duration)
    @name     = name
    @company   = company
    @duration = duration
  end
end

aSong1 = Employee.new("Fleck", "AMZ", 260)
aSong2 = Employee.new("Taylor", "EMC", 120)
aSong3 = Employee.new("Bob", "Adobe", 260)
aSong4 = Employee.new("Jack", "Google", 360)

final_array = [ ]
final_array.push(aSong1)
final_array.push(aSong2)
final_array.push(aSong3)
final_array.push(aSong4)

控制器

puts final_array.length #4
final_array.each do | element | 
  puts element.is_a?(Object) #true
  puts element.name #prints name
end

结果数组（预期）

result = [{company: 'AMZ',duration: 260}, {company: 'EMC',duration: 120},{company: 'Adobe',duration: 260}, {company: 'Google',duration: 360}  ]

示例：repl

Answer 1

您可以使用

删除ID键

 a = [{"id":"21","company":"AMC","name":"Matt"},{"id":"22","company":"AMC","name":"Jon"},
{"id":"12","company":"XYZ","name":"Bob"}].each{|o| o.delete :id}

如果你想将它与其他哈希进行比较，你可以使用merge方法，假设a2是第二个哈希

a.merge(a2)

这将返回一个新的哈希值，其中任何匹配的键都会使用第二个哈希值中的新对应值进行更新

Answer 2

final_array.collect {|x| { company: x.company, duration: x.duration }}

结果：

[{:company=>"AMZ", :duration=>260}, {:company=>"EMC", :duration=>120}, {:company=>"Adobe", :duration=>260}, {:company=>"Google", :duration=>360}]

Answer 3

以下情况如何？

class Employee
  def to_h
    instance_variables.each_with_object({}) do |var, h|
      k = var.to_s.tr('@','').to_sym
      v = instance_variable_get(var)
      h[k] = v
    end
  end
end

result = final_array.map{|e| e.to_h.delete_if{|k,v| k == :name}}

从对象数组中删除元素并创建哈希数组

示例创建

控制器

结果数组（预期）

3 个答案: