定义元素离XML根的距离

Question

考虑XML：

<items>
    <item id="0001" type="donut">
        <name>Cake</name>
        <ppu>0.55</ppu>
        <batters>
            <batter id="1001">Regular</batter>
            <batter id="1002">Chocolate</batter>
            <batter id="1003">Blueberry</batter>
        </batters>
        <topping id="5001">None</topping>
        <topping id="5002">Glazed</topping>
        <topping id="5005">Sugar</topping>
        <topping id="5006">Sprinkles</topping>
        <topping id="5003">Chocolate</topping>
        <topping id="5004">Maple</topping>
    </item>
</items>

项目是根，所以距离== 0

项目直接在根目录下，因此距离为1

名称是2个等级，所以距离将是2

如何在C＃中为XElement动态定义这样的距离？

Answer 1

您可以从System.Xml.XmlTextReader.Depth简化MSDN示例以显示节点元素及其各自的深度：
// XML file to be parsed.
string xmlFilePath = @"C:\test.xml";

// Create the reader.
using XmlTextReader reader = new XmlTextReader(xmlFilePath);

// Parse the XML and display each node.
while (reader.Read())
{
    // If node type is an element
    // Display element name and depth
    if (reader.NodeType == XmlNodeType.Element)
    {
        Console.WriteLine($"Element = {reader.Name}, Depth = {reader.Depth}");
    }
}
输出：
Element = items, Depth = 0
Element = item, Depth = 1
Element = name, Depth = 2
Element = ppu, Depth = 2
Element = batters, Depth = 2
Element = batter, Depth = 3
Element = batter, Depth = 3
Element = batter, Depth = 3
Element = topping, Depth = 2
Element = topping, Depth = 2
Element = topping, Depth = 2
Element = topping, Depth = 2
Element = topping, Depth = 2
Element = topping, Depth = 2

Answer 2

Answer 3

[使用任何方法选择目标元素，例如，id：//*[@id="1001"]。

选择所有祖先：//*[@id="1001"]/ancestor::*

计算那些祖先：count(//*[@id="1001"]/ancestor::*)。