Re: [xsl] xslt document function - recursive folder processing.

[Abel Braaksma <abel.online@xxxxxxxxx>]

> This is my first post;

Welcome!

> How can I keep track of the folder attribute in the parent template 

You're challenge is that you start from a new document node, from that 
point on, the original document node is out of scope. However, 
xsl:apply-templates can have a child, xsl:with-param. Use 
<xsl:with-param name="filename" select="@name" /> and inside the 
matching template, add <xsl:param name="filename" />. You can then refer 
to this by using the variable $filename.

> The potion of code below fails until I explictly provide the absolute
> path and also prepend it with file://D|

Not sure what the question is for Q #2, but your conclusion is correct. 
You need to provide a URI. The exact syntax for a file-uri is often 
debated and can be processor dependent (i.e., some may require file:///, 
as after file:// for protocol, you must first provide the root /).

> Is it possible for me to determine using XSLT the directory path of
> the XML file being processed?

Certain processors may provide extension functions for that. In XSLT 
2.0, you can use the base-uri() function to find the filename (not 
quite: to find the base uri) of a document. You can even use 
base-uri(documen('')) to get the filename of the executing XSLT file. If 
you need to stick to XSLT 1.0, you can pass the filename as a parameter 
on the commandline. Define the parameter as xsl:param in the root of 
your styleseet. Same solution works for getting the absolute current 
path (you Q #2).

Cheers,
Abel





------------------------------------------------------------------------
From: 	Lighton Phiri <lighton.phiri@xxxxxxxxx>
Sent: 	Saturday, August 06, 2011 12:34:26 PM
To: 	xsl-list
Cc: 	
Subject: 	[xsl] xslt document function - recursive folder processing.



> I am working with XSLT version 1.0 using xsltproc.
>
> This is my first post; please bare with me if I've missed out on any
> of the house rules. I have tried as much as possible to replicate what
> I am trying to do below. My senario is as follows:
>
> Folder Structure
>
> folder/
> |-- index.xml
> |-- dir1
> |   |-- d1f1.xml
> |   |-- d1f2.xml
> |   |--
> |-- dir2
> |   |-- d2f1.xml
> |   |-- d2f2.xml
> |   |--
>
>
> Current XSLT stylesheet
>
>   <?xml version="1.0" encoding="utf-8"?>
>
>     <xsl:stylesheet version="1.0"
>     xmlns:xsl="http://www.w3.org/1999/XSL/Transform";>
>
>       <xsl:output method="xml" indent="yes" encoding="utf-8" />
>
>       <xsl:template match="folder">
>         <xsl:apply-templates select="document(file)/file" />
>       </xsl:template>
>
>       <xsl:template match="file">
>         <name>
>       <xsl:value-of select="name" />
>         </name>
>         <link>
>       <file>document(../dir2/d2f1.xml, /)/file/name</file>
>       <description>document(../dir2/d2f1.xml, /)/file/description</description>
>         </link>
>       </xsl:template>
>
>     </xsl:stylesheet>
>
>
> index.xml file
>
> <?xml version="1.0" encoding="utf-8"?>
> <folder name="dir1">
>    <file>f1.xml</file>
>    <file>f2.xml</file>
> </folder>
>
>
> d1f1.xml
>
> <?xml version="1.0" encoding="utf-8"?>
> <file>
>    <name>File 1</name>
>    <description>I am file 1</description>
>    <link>dir2/d2f1.xml</link>
> </file>
>
> d2f2.xml
>
> <?xml version="1.0" encoding="utf-8"?>
> <file>
>    <name>File 2</name>
>    <description>I am file 2</description>
> </file>
>
>
> Question# 1:
> How can I keep track of the folder attribute in the parent template so
> I can make use of it when calling the child templates?
>
> Question# 2:
>
> The potion of code below fails until I explictly provide the absolute
> path and also prepend it with file://D|
>
> this fails
>         <link>
>       <file>document(../dir2/d2f1.xml, /)/file/name</file>
>       <description>document(../dir2/d2f1.xml, /)/file/description</description>
>         </link>
>
> this works (I stole this idea from
> http://www.dpawson.co.uk/xsl/sect2/N2602.html#d3862e403)
>
>         <link>
>       <file>document(file://D|/home/phiri/dir2/d2f1.xml)/file/name</file>
>       <description>document(file://D|/home/phiri/dir2/d2f1.xml)/file/description</description>
>         </link>
>
> Question# 3:
> Is it possible for me to determine using XSLT the directory path of
> the XML file being processed?
> All sources I have found online seem to suggest its not possible to
> evaluate the directory path of the XML file being processed.