Question

现在我的rails应用程序中有这样的数据：

[[{:car=>"", :article=>"", :group=>"", :price=>""},
  {:car=>"Volkswagen", :article=>"7H0127401A", :group=>"FILTER", :price=>"0,564"},
  {:car=>"Volkswagen", :article=>"7H0127401B", :group=>"FILTER", :price=>"0,546"}, 
  {:car=>"Volkswagen", :article=>"7H0127401D", :group=>"FILTER", :price=>"0,42"}], 
 [{:car=>"", :article=>"", :group=>"", :price=>""}, 
  {:car=>"Volkswagen", :article=>"7H0127401A", :group=>"FILTER", :price=>"0,564"}],   
 [{:car=>"", :article=>"", :group=>"", :price=>""}, 
  {:car=>"Volkswagen", :article=>"7H0127401B", :group=>"FILTER", :price=>"0,546"}]]

并查看：

=@oem_art.each do |oem|
      -oem.each do |oo|
        %tr{:class => cycle("zebra-stripe zebra-grey zebra1", "zebra-stripe zebra-grey zebra2")}   
          %td
            = oo[:car]
          %td
            %h4
              = oo[:article]            
          %td
            = oo[:group]
          %td
            = oo[:price]

但我怎么能以更美丽的方式做到这一点？另外如何使用空键跳过哈希？我怎么能通过标准的一个循环来做，例如：

=@oem_art.each do |oem|
          = oem.car
          %td
            =oem.price

怎么样呢？但这里没有用完......

UPD

  def original_art
    @article = Article.find_by_ART_ID(params[:id])
    @constr_num = ArtLookup.get_construction_number(@article.ART_ID)
    oem_art = []
    @constr_num.each do |o|
      as_oem = get_from_as_oem(o.ARL_SEARCH_NUMBER)
      if as_oem.present?
        oem_art << as_oem
      end
    end    
    @oem_art = oem_art.to_a.uniq
    @article = Article.first
    @oem_art
  end

Answer 1

美是个人品味的问题，flatten有助于避免嵌套循环。

- @oem_art.flatten.reject { |h| h[:car].empty? }.each do |oem|
  = oem[:car]
  # And so on ...

Answer 2

这是我在问题评论中提出的观点。如果您有oem_art：

[[{:car=>"", :article=>"", :group=>"", :price=>""},
  {:car=>"Volkswagen", :article=>"7H0127401A", :group=>"FILTER", :price=>"0,564"},
  {:car=>"Volkswagen", :article=>"7H0127401B", :group=>"FILTER", :price=>"0,546"}, 
  {:car=>"Volkswagen", :article=>"7H0127401D", :group=>"FILTER", :price=>"0,42"}], 
 [{:car=>"", :article=>"", :group=>"", :price=>""}, 
  {:car=>"Volkswagen", :article=>"7H0127401A", :group=>"FILTER", :price=>"0,564"}],   
 [{:car=>"", :article=>"", :group=>"", :price=>""}, 
  {:car=>"Volkswagen", :article=>"7H0127401B", :group=>"FILTER", :price=>"0,546"}]]

即。要素是：

1. [{:car=>"", :article=>"", :group=>"", :price=>""},
      {:car=>"Volkswagen", :article=>"7H0127401A", :group=>"FILTER", :price=>"0,564"},
      {:car=>"Volkswagen", :article=>"7H0127401B", :group=>"FILTER", :price=>"0,546"}, 
      {:car=>"Volkswagen", :article=>"7H0127401D", :group=>"FILTER", :price=>"0,42"}]

2. [{:car=>"", :article=>"", :group=>"", :price=>""}, 
      {:car=>"Volkswagen", :article=>"7H0127401A", :group=>"FILTER", :price=>"0,564"}]

3. [{:car=>"", :article=>"", :group=>"", :price=>""}, 
      {:car=>"Volkswagen", :article=>"7H0127401B", :group=>"FILTER", :price=>"0,546"}]

如果你致电oem_art.uniq，这将删除重复的元素，但3个元素都是不同的，所以不会删除任何内容。它实际上不会从元素中删除7H0127401A和7H0127401B的重复项。

Rails查看哈希数组

2 个答案: