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Question
Thursday, May 12, 2011 8:32 PM
Hi i am using the following code to query my XML file
var query = from item in xml.Root.Element("CLASS_A").Descendants("Something")
select item;
foreach (var record in query)
{
XElement my = record.Parent;
Console.WriteLine("");
Console.WriteLine("Parent name:{0}", my.Parent.Name);
Console.WriteLine("Location of record in XML :{0}", ????);> Need this
Console.WriteLine("");
}
How do i get the path of the record retrieved above so i could use it as an XPATH below
XmlDocument xml = new XmlDocument();
xml.LoadXml(str);
XmlNodeList xnList = xml.SelectNodes("Need the above record Path") << I need the recored path from above so i could use it here
Any help or suggestion would be appreciated thanks
A candle loses nothing by lighting another candle.
All replies (9)
Friday, May 13, 2011 7:30 AM ✅Answered
Hello,
using System;
using System.Text;
using System.Xml;
namespace XmlDocClassA
{
class Program
{
static void Main(string[] args)
{
const string str = @"<root>
<CLASS_A>
<Something></Something>
<someNode>
<Something></Something>
</someNode>
</CLASS_A>
</root>";
XmlDocument xmldoc = new XmlDocument();
xmldoc.LoadXml(str);
XmlNodeList list = xmldoc.SelectNodes("//CLASS_A//Something");
foreach (XmlNode record in list)
{
XmlNode myParent = record.SelectSingleNode("../..");
Console.WriteLine("");
if (myParent != null)
Console.WriteLine("Parent name : {0}", myParent.Name);
Console.WriteLine("Location of record in XML : {0}", FindXPath(record)); //> Need this
Console.WriteLine("");
}
Console.ReadKey();
}
static string FindXPath(XmlNode node)
{
StringBuilder builder = new StringBuilder();
while (node != null)
{
switch (node.NodeType)
{
case XmlNodeType.Attribute:
builder.Insert(0, "/@" + node.Name);
node = ((XmlAttribute)node).OwnerElement;
break;
case XmlNodeType.Element:
int index = FindElementIndex((XmlElement)node);
builder.Insert(0, "/" + node.Name + "[" + index + "]");
node = node.ParentNode;
break;
case XmlNodeType.Document:
return builder.ToString();
default:
throw new ArgumentException("Only elements and attributes are supported");
}
}
throw new ArgumentException("Node was not in a document");
}
static int FindElementIndex(XmlElement element)
{
XmlNode parentNode = element.ParentNode;
if (parentNode is XmlDocument)
{
return 1;
}
XmlElement parent = (XmlElement)parentNode;
int index = 1;
foreach (XmlNode candidate in parent.ChildNodes)
{
if (candidate is XmlElement && candidate.Name == element.Name)
{
if (candidate == element)
{
return index;
}
index++;
}
}
throw new ArgumentException("Couldn't find element within parent");
}
}
}
Kind regards,
Tuesday, May 17, 2011 9:11 PM ✅Answered
The xpath returned of your function does not contain attributes. It may be a problem if a node is a child of two nodes having the same name but different attribute values.
This function provides a valid xpath :
static string FindXPath(XmlNode node)
{
StringBuilder builder = new StringBuilder();
while (node != null)
{
switch (node.NodeType)
{
case XmlNodeType.Attribute:
builder.Insert(0, "/@" + node.Name);
node = ((XmlAttribute)node).OwnerElement;
break;
case XmlNodeType.Element:
int index = FindElementIndex((XmlElement)node);
builder.Insert(0, "/" + node.Name + "[" + index + "]");
node = node.ParentNode;
break;
case XmlNodeType.Document:
return builder.ToString();
default:
throw new ArgumentException("Only elements and attributes are supported");
}
}
throw new ArgumentException("Node was not in a document");
}
static int FindElementIndex(XmlElement element)
{
XmlNode parentNode = element.ParentNode;
if (parentNode is XmlDocument)
{
return 1;
}
XmlElement parent = (XmlElement)parentNode;
int index = 1;
foreach (XmlNode candidate in parent.ChildNodes)
{
if (candidate is XmlElement && candidate.Name == element.Name)
{
if (candidate == element)
{
return index;
}
index++;
}
}
throw new ArgumentException("Couldn't find element within parent");
}
Thursday, May 12, 2011 9:37 PM
Hi,
Firs you will need to retrieve all your Something nodes, XPATH :
/*/CLASS_A/*//Something
Then in a foreach loop, you can retrieve for each Something node its parent, XPATH :
..
Kind regards,
Thursday, May 12, 2011 10:24 PM
Hi,
Firs you will need to retrieve all your Something nodes, XPATH :
/*/CLASS_A/*//SomethingThen in a foreach loop, you can retrieve for each Something node its parent, XPATH :
..Kind regards,
I am not really sure what that means (sorry abt that) , could you kindly give a small code sample .. ??
A candle loses nothing by lighting another candle.
Monday, May 16, 2011 7:37 AM
Check this tutorial for more about XPath:
http://www.w3schools.com/XPath/default.asp Leo Liu [MSFT]
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Monday, May 16, 2011 4:57 PM
static string XPath(XElement e)
{
if (e == null)
return null;
else
return XPath(e.Parent) + "/" + e.Name;
}
Monday, May 16, 2011 5:09 PM
What about attributes and node index ?
aelassas.free.fr
Tuesday, May 17, 2011 4:55 PM
What is the question?
Wednesday, May 18, 2011 12:46 PM
You're right. Indexes are good.