获取当前节点的完整路径

Question

如果我XPathNavigator在一个节点上有一个定位,我如何从根目录获得一个表示该节点路径的XPath表达式？

例如,如果XML是:

<data>
    <class name='dogs'>
        <item name='doberman />
        <item name='husky' />
    </class>
    <class name='cats'>
        <item name='persian' />
        <item name='tabby' />
    </class> </data>
</data>

......那么波斯猫的路径可以表示为 /data/class[2]/item[1]

我可以枚举相关节点的祖先SelectAncestors()(或者我可以迭代地爬上父亲关系SelectParent()),但这并不能获得位置信息.

我是否必须position()为每个祖先评估使用XPath ,还是有更好的方法来执行此操作？

Answer 1

假设你只对xml元素的xpath感兴趣,我实现了一个强力算法(即遍历XML结构)作为扩展方法XmlElement.这与@ Zenexer的回答非常相似,尽管我在发布他的时候已经开始使用我自己的版本了.

另外,对于Alexei关于性能的提示感兴趣,我使用一个复杂的XML文件创建了一种测试用例.然后我实现了相同算法的两个版本; 一个依赖于PreviousSibling,另一个依赖迭代节点.第三个版本依赖于XPath的position()功能,但它没有按预期工作并被丢弃.

虽然您应该检查自己,但在我的机器中,结果显示迭代版本具有显着的性能优势 - 兄弟姐妹版本对21s进行了1.7s.

Importart:这些扩展方法在a中声明static class XmlElementExtension.

PreviousSibling版本

    public static string GetXPath_UsingPreviousSiblings(this XmlElement element)
    {
        string path = "/" + element.Name;

        XmlElement parentElement = element.ParentNode as XmlElement;
        if (parentElement != null)
        {
            // Gets the position within the parent element, based on previous siblings of the same name.
            // However, this position is irrelevant if the element is unique under its parent:
            XPathNavigator navigator = parentElement.CreateNavigator();
            int count = Convert.ToInt32(navigator.Evaluate("count(" + element.Name + ")"));
            if (count > 1) // There's more than 1 element with the same name
            {
                int position = 1;
                XmlElement previousSibling = element.PreviousSibling as XmlElement;
                while (previousSibling != null)
                {
                    if (previousSibling.Name == element.Name)
                        position++;

                    previousSibling = previousSibling.PreviousSibling as XmlElement;
                }

                path = path + "[" + position + "]";
            }

            // Climbing up to the parent elements:
            path = parentElement.GetXPath_UsingPreviousSiblings() + path;
        }

        return path;
    }

迭代版

    public static string GetXPath_SequentialIteration(this XmlElement element)
    {
        string path = "/" + element.Name;

        XmlElement parentElement = element.ParentNode as XmlElement;
        if (parentElement != null)
        {
            // Gets the position within the parent element.
            // However, this position is irrelevant if the element is unique under its parent:
            XmlNodeList siblings = parentElement.SelectNodes(element.Name);
            if (siblings != null && siblings.Count > 1) // There's more than 1 element with the same name
            {
                int position = 1;
                foreach (XmlElement sibling in siblings)
                {
                    if (sibling == element)
                        break;

                    position++;
                }

                path = path + "[" + position + "]";
            }

            // Climbing up to the parent elements:
            path = parentElement.GetXPath_SequentialIteration() + path;
        }

        return path;
    }

测试用例

    private static void Measure(string functionName, int iterations, Action implementation)
    {
        Stopwatch watch = new Stopwatch();
        watch.Start();

        for (int i = 0; i < iterations; i++)
        {
            implementation();
        }

        watch.Stop();
        Console.WriteLine("{0}: {1}ms", functionName, watch.ElapsedMilliseconds);
    }

    private static void Main(string[] args)
    {
        XmlDocument doc = new XmlDocument();
        doc.Load(@"location of some large and complex XML file");

        string referenceXPath = "/vps/vendorProductSets/vendorProductSet/product[100]/prodName/locName";

        Measure("UsingPreviousSiblings", 10000,
                () =>
                    {
                        XmlElement target = doc.SelectSingleNode(referenceXPath) as XmlElement;
                        Debug.Assert(referenceXPath == target.GetXPath_UsingPreviousSiblings());
                    });

        Measure("SequentialIteration", 10000,
                () =>
                {
                    XmlElement target = doc.SelectSingleNode(referenceXPath) as XmlElement;
                    Debug.Assert(referenceXPath == target.GetXPath_SequentialIteration());
                });
    }