Как получить xpath из экземпляра XmlNode

Question

Может ли кто-нибудь предоставить некоторый код, который получит xpath экземпляра System.Xml.XmlNode?

Спасибо!

Answer 1

Если вы сделаете это, вы получите Путь с Именами Узлов И Позицией, если у вас есть Узлы с таким же именем, например: "/ Услуги [1] / System [1] / Группа [1] / Папка [2] / File [2]"

public string GetXPathToNode(XmlNode node)
{         
    if (node.NodeType == XmlNodeType.Attribute)
    {             
        // attributes have an OwnerElement, not a ParentNode; also they have             
        // to be matched by name, not found by position             
        return String.Format("{0}/@{1}", GetXPathToNode(((XmlAttribute)node).OwnerElement), node.Name);
    }
    if (node.ParentNode == null)
    {             
        // the only node with no parent is the root node, which has no path
        return "";
    }

    //get the index
    int iIndex = 1;
    XmlNode xnIndex = node;
    while (xnIndex.PreviousSibling != null && xnIndex.PreviousSibling.Name == xnIndex.Name)
    {
         iIndex++;
         xnIndex = xnIndex.PreviousSibling; 
    }

    // the path to a node is the path to its parent, plus "/node()[n]", where
    // n is its position among its siblings.         
    return String.Format("{0}/{1}[{2}]", GetXPathToNode(node.ParentNode), node.Name, iIndex);
}

Answer 2

Я должен был сделать это недавно. Только элементы должны быть рассмотрены. Вот что я придумал:

    private string GetPath(XmlElement el)
    {
        List<string> pathList = new List<string>();
        XmlNode node = el;
        while (node is XmlElement)
        {
            pathList.Add(node.Name);
            node = node.ParentNode;
        }
        pathList.Reverse();
        string[] nodeNames = pathList.ToArray();
        return String.Join("/", nodeNames);
    }

Answer 3

 public static string GetFullPath(this XmlNode node)
        {
            if (node.ParentNode == null)
            {
                return "";
            }
            else
            {
                return $"{GetFullPath(node.ParentNode)}\\{node.ParentNode.Name}";
            }
        }

Answer 4

Это еще проще

 ''' <summary>
    ''' Gets the full XPath of a single node.
    ''' </summary>
    ''' <param name="node"></param>
    ''' <returns></returns>
    ''' <remarks></remarks>
    Private Function GetXPath(ByVal node As Xml.XmlNode) As String
        Dim temp As String
        Dim sibling As Xml.XmlNode
        Dim previousSiblings As Integer = 1

        'I dont want to know that it was a generic document
        If node.Name = "#document" Then Return ""

        'Prime it
        sibling = node.PreviousSibling
        'Perculate up getting the count of all of this node's sibling before it.
        While sibling IsNot Nothing
            'Only count if the sibling has the same name as this node
            If sibling.Name = node.Name Then
                previousSiblings += 1
            End If
            sibling = sibling.PreviousSibling
        End While

        'Mark this node's index, if it has one
        ' Also mark the index to 1 or the default if it does have a sibling just no previous.
        temp = node.Name + IIf(previousSiblings > 0 OrElse node.NextSibling IsNot Nothing, "[" + previousSiblings.ToString() + "]", "").ToString()

        If node.ParentNode IsNot Nothing Then
            Return GetXPath(node.ParentNode) + "/" + temp
        End If

        Return temp
    End Function