Find position of a node using XPath - xpath

Question

Anyone know how to get the position of a node using XPath?
Say I have the following xml:
<a>
<b>zyx</b>
<b>wvu</b>
<b>tsr</b>
<b>qpo</b>
</a>
I can use the following xpath query to select the third <b> node (<b>tsr</b>):
a/b[.='tsr']
Which is all well and good but I want to return the ordinal position of that node, something like:
a/b[.='tsr']/position()
(but a bit more working!)
Is it even possible?
edit: Forgot to mention am using .net 2 so it's xpath 1.0!
Update: Ended up using James Sulak's excellent answer. For those that are interested here's my implementation in C#:
int position = doc.SelectNodes("a/b[.='tsr']/preceding-sibling::b").Count + 1;
// Check the node actually exists
if (position > 1 || doc.SelectSingleNode("a/b[.='tsr']") != null)
{
Console.WriteLine("Found at position = {0}", position);
}

Answer 1

Try:
count(a/b[.='tsr']/preceding-sibling::*)+1.

Answer 2

You can do this with XSLT but I'm not sure about straight XPath.
<?xml version="1.0" encoding="UTF-8" standalone="yes"?>
<xsl:stylesheet version="1.0" xmlns:xsl="http://www.w3.org/1999/XSL/Transform">
<xsl:output method="xml" encoding="utf-8" indent="yes"
omit-xml-declaration="yes"/>
<xsl:template match="a/*[text()='tsr']">
<xsl:number value-of="position()"/>
</xsl:template>
<xsl:template match="text()"/>
</xsl:stylesheet>

Answer 3

I realize that the post is ancient.. but..
replace'ing the asterisk with the nodename would give you better results
count(a/b[.='tsr']/preceding::a)+1.
instead of
count(a/b[.='tsr']/preceding::*)+1.

Answer 4

If you ever upgrade to XPath 2.0, note that it provides function index-of, it solves problem this way:
index-of(//b, //b[.='tsr'])
Where:
1st parameter is sequence for searching
2nd is what to search

Answer 5

Unlike stated previously 'preceding-sibling' is really the axis to use, not 'preceding' which does something completely different, it selects everything in the document that is before the start tag of the current node. (see http://www.w3schools.com/xpath/xpath_axes.asp)

Answer 6

Just a note to the answer done by James Sulak.
If you want to take into consideration that the node may not exist and want to keep it purely XPATH, then try the following that will return 0 if the node does not exist.
count(a/b[.='tsr']/preceding-sibling::*)+number(boolean(a/b[.='tsr']))

Answer 7

The problem is that the position of the node doesn't mean much without a context.
The following code will give you the location of the node in its parent child nodes
using System;
using System.Xml;
public class XpathFinder
{
public static void Main(string[] args)
{
XmlDocument xmldoc = new XmlDocument();
xmldoc.Load(args[0]);
foreach ( XmlNode xn in xmldoc.SelectNodes(args[1]) )
{
for (int i = 0; i < xn.ParentNode.ChildNodes.Count; i++)
{
if ( xn.ParentNode.ChildNodes[i].Equals( xn ) )
{
Console.Out.WriteLine( i );
break;
}
}
}
}
}

Answer 8

I do a lot of Novell Identity Manager stuff, and XPATH in that context looks a little different.
Assume the value you are looking for is in a string variable, called TARGET, then the XPATH would be:
count(attr/value[.='$TARGET']/preceding-sibling::*)+1
Additionally it was pointed out that to save a few characters of space, the following would work as well:
count(attr/value[.='$TARGET']/preceding::*) + 1
I also posted a prettier version of this at Novell's Cool Solutions: Using XPATH to get the position node