ruby – 从物化路径构建树

发布时间：2020-12-17 02:30:25 所属栏目：百科来源：网络整理

导读：我在使用 ruby从物化路径构建树结构时遇到了麻烦. 假设我有一个排序结果集(来自couchdb)： [ { :key = [],:value = "Home" },{ :key = ["about"],:value = "About" },{ :key = ["services"],:value = "Services" },{ :key = ["services","plans"],:value = "

我在使用 ruby从物化路径构建树结构时遇到了麻烦.

假设我有一个排序结果集(来自couchdb)：

[
  { :key => [],:value => "Home" },{ :key => ["about"],:value => "About" },{ :key => ["services"],:value => "Services" },{ :key => ["services","plans"],:value => "Plans" },"training"],:value => "Training" },"training","python"],:value => "Python" },"ruby"],:value => "Ruby" }
]

我只需要将它作为ruby中的树,以下哈希就足够了：

{ :title => "Home",:path => [],:children => [
  { :title => "About",:path => ["about"] },{ :title => "Services",:path => ["services"],:children => [
    { :title => "Plans",:path => ["services","plans"] }
  ]}
]}

谁能帮助我？

解决方法

您需要一个简单的帮助类和一点递归：

class Tree
  attr_reader :root

  def initialize
    @root = { :title => 'Home',:path => [ ],:children => [ ] }
  end

  def add(p)
    r_add(@root,p[:key].dup,p[:value])
    self
  end

private

  def r_add(h,path,value)
    if(path.empty?)
      h[:title] = value 
      return
    end

    p = path.shift
    c = h[:children].find { |c| c[:path].last == p } 
    if(!c)
      c = { :title => nil,:path => h[:path].dup.push(p),:children => [ ] }
      h[:children].push(c)
    end
    r_add(c,value)
  end

end

然后：

t = a.inject(Tree.new) { |t,h| t.add(h) }
h = t.root

会在h中给出这个：

{:title =>"Home",:path=>[],:children=>[
  {:title=>"About",:path=>["about"],:children=>[]},{:title=>"Services",:path=>["services"],:children=>[
    {:title=>"Plans",:path=>["services",{:title=>"Training",:children=>[
      {:title=>"Python",{:title=>"Ruby",:children=>[]}
    ]}
  ]}
]}

你可以理清空：孩子,如果他们重要.

（编辑：李大同）

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