Hi Abel, thank you very much.
Question#1: How can I keep track of the folder attribute in the parent template
(a). If I get what you mean, then this would effectively only work
with the parent nodes accessing child parameters; what I want to do
however is to access parent attributes in child nodes.
(b). If that is possible, do you suppose it would be feasible to use
this approach for a 3/4-level hierarchy of folders?
Question# 2:
What I meant to say in question 2 is: Is it possible to use the
file://D| syntax with relative path names? It seems not to work and I
was thinking that perhaps I could have been doing something wrong; my
implementation requires the use of relative pathnames --for now.
On 6 August 2011 14:34, Abel Braaksma <abel(_dot_)online(_at_)xs4all(_dot_)nl>
wrote:
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(_dot_)phiri(_at_)gmail(_dot_)com>
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.
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