How to get line numbers when evaluating XPath-expressions?

[linuxadmin]

Hello everyone!

I evaluate XPath expressions using XPathNavigator and need to know for
each result, which line number (or et least file position) it has
inside the XML file.

Any ideas?

Thanks in advice!

 

[Guest]

The problem here is that An XPathNavigator is navigating an XPath Document
tree, which has only nodes, parents, children etc. and has no "concept" of a
"line number in a file". Make sense?

--Peter
"Inside every large program, there is a small program trying to get out."
http://www.eggheadcafe.com
http://petesbloggerama.blogspot.com
http://www.blogmetafinder.com

 

[linuxadmin]

Hello!

Isn't it possible to "locate" an XPathNavigator (XPathNode?) after it
was returned, using XmlReader or something else?

 

[Guest]

Yes, it is possible but quite involved.

Your task is a lot easier if you know that each element, comment &
processing instruction starts on a new line and that there are no empty
lines. If this isn't the case you may want to create your own version of the
XML file first, preserving whitespace is an issue here though.

XmlReader can be used but it does not return its location within the file,
so you will need to keep track of this yourself, or you may want to use
TextReader with regular expressions.

The key thing you need is the ordinal position of the element immediately
preceding the returned node from the expression (this helps cover the case
when a node returned is a mixed-content text node) using an xpath expression.

SketchPath is an example XPath application that uses this technique on its
own representation of the XML file.

The following rough code snippet (from SketchPath) might give your some
ideas, this is incomplete, only caters for element nodes and doesn't take
account of comments and processing-instructions.


XPathExpression expr = xnav.Compile(contextXpath);
expr.SetContext(namespaceData);
if (expr.ReturnType == XPathResultType.NodeSet)
{
XPathNodeIterator iterator =
(XPathNodeIterator)xnav.Evaluate(expr);
iterator.MoveNext();
if (iterator.Current.NodeType == XPathNodeType.Element)
{
xnav = (XPathNavigator)iterator.Current.Clone(); //
preserve location of context element in navigator

// find number of preceding elements
double elCount =
(double)xnav.Evaluate("count(./preceding::*) + count(./ancestor::*)");
....

Phil Fearon
http://www.sketchpath.com

 

[linuxadmin]

Thanks a lot!
I'll try it out

Yes, it is possible but quite involved.

Your task is a lot easier if you know that each element, comment &
processing instruction starts on a new line and that there are no empty
lines. If this isn't the case you may want to create your own version of the
XML file first, preserving whitespace is an issue here though.

XmlReader can be used but it does not return its location within the file,
so you will need to keep track of this yourself, or you may want to use
TextReader with regular expressions.

The key thing you need is the ordinal position of the element immediately
preceding the returned node from the expression (this helps cover the case
when a node returned is a mixed-content text node) using an xpath expression.

SketchPath is an example XPath application that uses this technique on its
own representation of the XML file.

The following rough code snippet (from SketchPath) might give your some
ideas, this is incomplete, only caters for element nodes and doesn't take
account of comments and processing-instructions.

XPathExpression expr = xnav.Compile(contextXpath);
expr.SetContext(namespaceData);
if (expr.ReturnType == XPathResultType.NodeSet)
{
XPathNodeIterator iterator =
(XPathNodeIterator)xnav.Evaluate(expr);
iterator.MoveNext();
if (iterator.Current.NodeType == XPathNodeType.Element)
{
xnav = (XPathNavigator)iterator.Current.Clone(); //
preserve location of context element in navigator

// find number of preceding elements
double elCount =
(double)xnav.Evaluate("count(./preceding::*) + count(./ancestor::*)");
....

Phil Fearonhttp://www.sketchpath.com