Question

我有一个我传递的参数，看起来像这样：

{{1}}

如何迭代此参数以获取每个项目的值？我知道我可以单独抓取每个项目，但在提交动态表单数据时，我不知道参数中的项目的确切数量。我怎样才能解决这个问题？

Answer 1

hash = {"item_0"=>{"0"=>["address_1"], "1"=>["model 1"], "2"=>["ABC"], "3"=>["Off"], "4"=>["Status"]}, "item_1"=>{"0"=>["address_2"], "1"=>["model 1"], "2"=>["DEF"], "3"=>["On"], "4"=>["Status"]}, "item_2"=>{"0"=>["address_3"], "1"=>["model 1"], "2"=>["GHI"], "3"=>["On"], "4"=>["Status"]}}

hash.each do |k1, h1|
  #eg: k1 = "item_0"; h1 = {"0"=>["address_1"], "1"=>["model 1"], "2"=>["ABC"], "3"=>["Off"], "4"=>["Status"]}
  h1.each do |k2,arr|
    #eg: k2 = "0"; arr = ["address_1"]
    val = arr.first
    puts "do something with k1 = #{k1.inspect}, k2 = #{k2.inspect}, val = #{val.inspect} here"
  end
end

我使用类似的模式来请求更新多个对象，我通过这种结构发送params：

params = {:resources => {123 => {:name => "Foo", :address => "bar"}, 456 => {:name => "Boo", :address => "Far"}}}

然后在我的代码中我可以做

params[:resources].each do |id, attrs|
  if resource = Resource.find_by_id(id)
    resource.update_attributes(attrs)
  end
end

可以用更长篇的方式写成

params[:resources].each do |id, attrs|
  if resource = Resource.find_by_id(id)
    attrs.each do |k,v|
      resource.send("#{k}=", v)
    end
    resource.save
  end
end

Answer 2

要仅保留您想要的密钥，您可以使用Hash#keep_if执行以下操作：

hsh = {"utf8"=>"✓",
       "authenticity_token"=>"xx==",
       "item_0"=>{"0"=>["address_1"], "1"=>["model 1"], "2"=>["ABC"], "3"=>["Off"], "4"=>["Status"]},
       "item_1"=>{"0"=>["address_2"], "1"=>["model 1"], "2"=>["DEF"], "3"=>["On"], "4"=>["Status"]},
       "item_2"=>{"0"=>["address_3"], "1"=>["model 1"], "2"=>["GHI"], "3"=>["On"], "4"=>["Status"]}}

hsh.keep_if {|k, v| k=~ /item_/ }

然后会返回：

{"item_0"=>{"0"=>["address_1"], "1"=>["model 1"], "2"=>["ABC"], "3"=>["Off"], "4"=>["Status"]}, 
"item_1"=>{"0"=>["address_2"], "1"=>["model 1"], "2"=>["DEF"], "3"=>["On"], "4"=>["Status"]},
"item_2"=>{"0"=>["address_3"], "1"=>["model 1"], "2"=>["GHI"], "3"=>["On"], "4"=>["Status"]}}

您可以使用Hash循环播放each_pair。每一对都参考例如代表键/值对的|k, v|。从那里你可以做你喜欢的哈希。由于你的哈希是嵌套的，我画了这个例子：

hsh = {"item_0"=>{"0"=>["address_1"], "1"=>["model 1"], "2"=>["ABC"],"3"=>["Off"], "4"=>["Status"]},
       "item_1"=>{"0"=>["address_2"], "1"=>["model 1"], "2"=>["DEF"], "3"=>["On"], "4"=>["Status"]},
       "item_2"=>{"0"=>["address_3"], "1"=>["model 1"], "2"=>["GHI"], "3"=>["On"], "4"=>["Status"]}}

> hsh.each_pair {|k, v| v.each_pair {|j, w| puts w } }
address_1
model 1
ABC
Off
Status
address_2
model 1
DEF
On
Status
address_3
model 1
GHI
On
Status

Here's指向Hash#each_pair的文档的链接。

有许多循环哈希的方法。例如，each_key和each_value，其功能与命名相同。

Rails - 通过Params迭代

2 个答案: