嵌套哈希和方法重用

Question

我有一些我想迭代的哈希值，还有一些嵌套的哈希值可以达到 3-4 层， 我目前正在使用 if 语句来检查该值是否为哈希值，然后再次遍历它，但我只是在这里重复代码。

有没有使用方法的 DRY 方法来做到这一点？

我还希望最终输出出现在表格中，最好的方法是什么？这就是为什么我最初有多个 if 语句，所以我可以添加单独的标签。

示例方法：

<% def hashTest(key, value) %>
    <% if value.is_a?(Hash) %>
        <%= key %>
            <% value.each do |key, value| %>
                <%= key %>
                <%= value %>
            <% end %>
    <% else %>
        <%= key %>
        <%= value %>
    <% end %>
<% end %>

这就是我的 if 语句的混乱......

<% parsed.each do |key, value| %>
    <% if value.is_a?(Hash) %>
        <%= key %>
        <br/>
        <% value.each do |key, value| %>
                <% if value.is_a?(Hash) %>
                    <%= key %>
                    <br/>
                    <% value.each do |key, value| %>
                            <% if value.is_a?(Hash) %>
                            <%= key %>
                            <br/>
                                <% value.each do |key, value| %>
                                    <%= key %>
                                    <%= value %>
                                    <br/>
                                <% end %>
                            <% else %>
                                <%= key %>
                                <%= value %>
                                <br/>
                            <% end %>
                    <% end %>
                <% else %>
                    <%= key %>
                    <%= value %>
                    <br/>
                <% end %>
        <% end %>
    <% else %>
        <%= key %>
        <%= value %>
        <br/>
    <% end %>
<% end %>

示例输入：

{
  "statement": {
    "generated": "2015-01-11",
    "due": "2015-01-25",
    "period": {
      "from": "2015-01-26",
      "to": "2015-02-25"
    }
  },
  "total": 136.03,
  "package": {
    "subscriptions": [
      { "type": "tv", "name": "Movies", "cost": 50.00 },
      { "type": "Phone", "name": "Landline", "cost": 5.00 },
      { "type": "broadband", "name": "Fibre", "cost": 16.40 }
    ],
    "total": 71.40
  },
  "callCharges": {
    "calls": [
      { "called": "07716393769", "duration": "00:23:03", "cost": 2.13 },
      { "called": "07716393769", "duration": "00:23:03", "cost": 2.13 },
      { "called": "07716393769", "duration": "00:23:03", "cost": 2.13 },
      { "called": "07716393769", "duration": "00:23:03", "cost": 2.13 },
      { "called": "07716393769", "duration": "00:23:03", "cost": 2.13 },
      { "called": "07716393769", "duration": "00:23:03", "cost": 2.13 },
      { "called": "07716393769", "duration": "00:23:03", "cost": 2.13 },
      { "called": "07716393769", "duration": "00:23:03", "cost": 2.13 },
      { "called": "07716393769", "duration": "00:23:03", "cost": 2.13 },
      { "called": "07716393769", "duration": "00:23:03", "cost": 2.13 },
      { "called": "07716393769", "duration": "00:23:03", "cost": 2.13 },
      { "called": "07716393769", "duration": "00:23:03", "cost": 2.13 },
      { "called": "07716393769", "duration": "00:23:03", "cost": 2.13 },
      { "called": "07716393769", "duration": "00:23:03", "cost": 2.13 },
      { "called": "07716393769", "duration": "00:23:03", "cost": 2.13 },
      { "called": "07716393769", "duration": "00:23:03", "cost": 2.13 },
      { "called": "07716393769", "duration": "00:23:03", "cost": 2.13 },
      { "called": "07716393769", "duration": "00:23:03", "cost": 2.13 },
      { "called": "02074351359", "duration": "00:23:03", "cost": 2.13 },
      { "called": "02074351359", "duration": "00:23:03", "cost": 2.13 },
      { "called": "02074351359", "duration": "00:23:03", "cost": 2.13 },
      { "called": "02074351359", "duration": "00:23:03", "cost": 2.13 },
      { "called": "02074351359", "duration": "00:23:03", "cost": 2.13 },
      { "called": "02074351359", "duration": "00:23:03", "cost": 2.13 },
      { "called": "02074351359", "duration": "00:23:03", "cost": 2.13 },
      { "called": "02074351359", "duration": "00:23:03", "cost": 2.13 },
      { "called": "02074351359", "duration": "00:23:03", "cost": 2.13 },
      { "called": "02074351359", "duration": "00:23:03", "cost": 2.13 }
    ],
    "total": 59.64
  },
  "Store": {
    "rentals": [
      { "title": "50 Shades of Grey", "cost": 4.99 }
    ],
    "purchases": [
      { "title": "That's what she said", "cost": 9.99 },
      { "title": "Broke back mountain", "cost": 9.99 }
    ],
    "total": 24.97
  }
}

示例输出：

https://jsfiddle.net/un5ex8y1/1/

Answer 1

假设您想在单独的行中显示每个键、每个值和每个数组元素，

def flatten_breakify(val)
  case val
  when Hash
    val.map { |k, v| "#{k}<br>#{flatten_breakify(v)}" }.join('<br>')
  when Array
    val.map(&method(:flatten_breakify)).join('<br>')
  else
    val
  end
end

另一种更优雅的方法：

def flatten_all(val)
  case val
  when Hash
    val.flat_map { |k, v| [k, *flatten_all(v)] }
  when Array
    val.flat_map(&method(:flatten_all))
  else
    val
  end
end

然后你可以放入你的模板

<%= flatten_all(data).join('<br>') %>

为了不将 HTML 混入控制器中。

编辑：我看到您在输出中添加了缩进，其他地方没有提到。我建议您使用 <ul> 而不是普通的 <br> 来实现正确的嵌套，并使用 CSS 解决缩进问题。您将无法使用第二种方法，因为它现在具有更多结构。第一段代码，因为它涉及 HTML，所以不应该在控制器中，而应该在帮助器中（这就是帮助器的用途，用于从数据生成 HTML）。