选择名称以某个字符串开头的所有XML元素?

I have two elements, item and date:

我有两个元素，项目和日期:

item has_many dates
项has_many日期
date belongs_to item
日期belongs_to项

I have a file with XML tree like this:

我有这样一个XML树文件:

<content>
  <item_1>
    <title>
    <description>
    <date_1>
      <date></date>
      <count></count>
    </date_1>
    <date_2>
      <date></date>
      <count></count>
    </date_2>
  </item_1>
  <item_2>
    <title>
    <description>
    <date_1>
      <date></date>
      <count></count>
    </date_1>
    <date_2>
      <date></date>
      <count></count>
    </date_2>
  </item_2>
</content>

I use Nokogiri for parsing data from item and date. I wrote a half of the script, which creates elements:

我使用Nokogiri解析项目和日期的数据。我写了剧本的一半，创造了元素:

doc.xpath("//content/*").each do |item|
  Item.create!(
              title: item.xpath("title").text,
              description: item.xpath("description").text)
end

Now I need find and build all dates of the items (element date), but can't understand how to parse tags like <data_1>, <data_2>, <data_3> etc. What I was trying

现在我需要找到并构建项目的所有日期(元素日期)，但是不理解如何解析标签，比如、、等等

  tour.xpath("//*/data_*").each do |date|
    puts date
  end

或

  tour.xpath{ |i| "//*/data_#{i}" }.each do |date|
    puts date
  end

but it doesn't work or I get an error.

但它不管用，否则我就会出错。

2 个解决方案

#1

The XPath for selecting all date elements in your document, regardless of their heritage, is simply:

选择文档中所有日期元素的XPath，不管它们的传统是什么，就是:

//date

The XPath for selecting all elements whose name starts with "date_" is:

选择名称以“date_”开头的所有元素的XPath是:

//*[starts-with(local-name(), 'date_')]

See starts-with() and local-name().

看到始于()和本地名称()。

#2

Try this:

试试这个:

doc.xpath("//content/*").each do |item|
  ...
  item.xpath("./*[starts-with(local-name(), 'date_')]").each do |d|
    ...
  end
end

(snipped edited according to @kjhughes suggestion)

(根据@kjhughes的建议进行剪辑)

#1