xsl-list Digest 1 Jul 2005 05:10:00 -0000 Issue 464
Topics (messages 18233 through 18282):
Re: sorting issue
18233 by: omprakash.v.polaris.co.in
18237 by: David Carlisle
Problems grouping adjacent siblings
18234 by: Huditsch Roman
18235 by: Mukul Gandhi
18238 by: David Carlisle
18239 by: Huditsch Roman
18260 by: Michael Kay
Re: problems with copying duplicate nodes
18236 by: Rahil Qamar
Re: [exslt] Re: [xsl] Importing compiled stylesheets
18240 by: James Fuller
18241 by: Colin Paul Adams
18277 by: Michael Kay
Cocoon and CSS
18242 by: Mike Holt
Re: [exslt] Re: Re: [xsl] Importing compiled stylesheets
18243 by: James Fuller
18273 by: Dimitre Novatchev
18276 by: Michael Kay
XSLT2/XPath2 books
18244 by: Sylvain Hellegouarch
18245 by: James Fuller
18246 by: Sylvain Hellegouarch
18247 by: John
18248 by: Sylvain Hellegouarch
18261 by: Michael Kay
18271 by: Dimitre Novatchev
18274 by: Sylvain Hellegouarch
Re: Spam:[xsl] sorting issue
18249 by: JBryant.s-s-t.com
priority attribute
18250 by: Arian Hojat
18251 by: Arian Hojat
18255 by: David Carlisle
18256 by: David Carlisle
Saxon SQL extension and Oracle Date format
18252 by: Perry Molendijk
18257 by: JBryant.s-s-t.com
an elements schema context via xslt
18253 by: Jan Limpens
18258 by: George Cristian Bina
18259 by: Arian Hojat
18266 by: Wendell Piez
18267 by: Jan Limpens
18269 by: Jan Limpens
18270 by: Wendell Piez
18278 by: Michael Kay
Re: priority attribute... Doh Solved
18254 by: Arian Hojat
Building OpenURL formatted URLs in XSLT
18262 by: GPoindexter.computer.org
18279 by: Michael Kay
Group problem with blank elements
18263 by: Ed Yau
18264 by: JBryant.s-s-t.com
18280 by: Michael Kay
18282 by: Mukul Gandhi
Re: test if null
18265 by: Wendell Piez
18275 by: Michael Kay
select all answers belongs to one question from xml with xpath
18268 by: 04083259.brookes.ac.uk
18272 by: Joris Gillis
XSL related activities at Extreme Markup Languages
18281 by: B. Tommie Usdin
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Date: Thu, 30 Jun 2005 11:48:33 +0530
To: xsl-list(_at_)lists(_dot_)mulberrytech(_dot_)com
From: omprakash(_dot_)v(_at_)polaris(_dot_)co(_dot_)in
Subject: Re: [xsl] sorting issue
Message-ID:
<OF1E8BFA1C(_dot_)F59486D6-ON65257030(_dot_)00221CE6(_at_)polaris(_dot_)co(_dot_)in>
Hi,
I have modified your stylesheet to make it work:
<?xml version="1.0"?>
<xsl:stylesheet version="1.0"
xmlns:xsl="http://www.w3.org/1999/XSL/Transform";>
<xsl:preserve-space elements="text"/>
<xsl:template match="masterfile">
<merged>
<table style="border:1px solid
#000000;border-collapse:collapse">
<xsl:apply-templates
select="document(doc/@filename)/client">
<!-- <xsl:sort select="name"/> -->
<!-- <xsl:sort select="subclient/name/lname" /> -->
</xsl:apply-templates>
</table>
</merged>
</xsl:template>
<xsl:template match="text()">
<xsl:value-of select="translate(., '

', ' ')"/>
</xsl:template>
<!-- newly added template for client and xsl:sort moved from above template
to here -->
<!-- major change was I modified the select attribute of sort to be
name/lname instead of subclient/name/lname -->
<xsl:template match="client">
<xsl:apply-templates select="subclient">
<xsl:sort select="name/lname" />
</xsl:apply-templates>
</xsl:template>
<xsl:template match="client/subclient">
<xsl:copy>
<!-- <tr>
<th>
<xsl:apply-templates select="title"/>
</th>
</tr> -->
<tr>
<td style="border:1px solid #000000">
<!-- <xsl:apply-templates select="@*|node()"/> -->
<!-- <xsl:apply-templates select="name"/> -->
<xsl:for-each select="name">
<xsl:sort select="lname" order="ascending"/>
<xsl:value-of select="fname"/>
<xsl:text> </xsl:text>
<xsl:value-of select="lname"/>
</xsl:for-each>
<!--
<xsl:apply-templates select="lname"/> -->
</td>
<td style="border:1px solid #000000">
<xsl:apply-templates select="phones"/>
</td>
<td style="border:1px solid #000000">
<xsl:apply-templates select="address"/>
</td>
</tr>
</xsl:copy>
</xsl:template>
<!-- <xsl:for-each select="name">
<xsl:sort select="lname"/>
<xsl:value-of select="fname"/>
</xsl:for-each> -->
<xsl:template match="address">
<xsl:for-each select="line">
<xsl:apply-templates/>
<br/>
</xsl:for-each>
<xsl:value-of select="url"/>
<br/>
<xsl:value-of select="email"/>
<br/>
</xsl:template>
<xsl:template match="name">
<xsl:apply-templates/>
<xsl:text> </xsl:text>
</xsl:template>
</xsl:stylesheet>
Hope this helps.
Cheers,
prakash
cookie king
<cookieking(_at_)comc To: xsl-list(_at_)lists(_dot_)mulberrytech(_dot_)com
ast.net> cc: (bcc: omprakash.v/Polaris)
Subject: [xsl] sorting issue
06/30/2005 06:27
AM
Please respond
to xsl-list
Recently submitted a post requesting help with sorting. my situation is
still the same. i am merging many documents. i get the sort to work but
not correctly. it appears to be looking at only the first node of each
file and sorting the files according to the first node. i need to sort
each individual file first. then merge the sort of all file, or sort
each file then merge all.
mastermerge.xsl
=====================================
<?xml version="1.0"?>
<xsl:stylesheet version="1.0"
xmlns:xsl="http://www.w3.org/1999/XSL/Transform";>
<xsl:preserve-space elements="text"/>
<xsl:template match="masterfile">
<merged>
<table style="border:1px solid
#000000;border-collapse:collapse">
<xsl:apply-templates
select="document(doc/@filename)/client">
<!-- <xsl:sort select="name"/> -->
<xsl:sort select="subclient/name/lname" />
</xsl:apply-templates>
</table>
</merged>
</xsl:template>
<xsl:template match="text()">
<xsl:value-of select="translate(., '

', ' ')"/>
</xsl:template>
<xsl:template match="client/subclient">
<xsl:copy>
<!-- <tr>
<th>
<xsl:apply-templates select="title"/>
</th>
</tr>
-->
<tr>
<td style="border:1px solid #000000">
<!-- <xsl:apply-templates select="@*|node()"/> -->
<!-- <xsl:apply-templates select="name"/> -->
<xsl:for-each select="name">
<xsl:sort select="lname" order="ascending"/>
<xsl:value-of select="fname"/>
<xsl:text> </xsl:text>
<xsl:value-of select="lname"/>
</xsl:for-each>
<!--
<xsl:apply-templates
select="lname"/> -->
</td>
<td style="border:1px solid #000000">
<xsl:apply-templates select="phones"/>
</td>
<td style="border:1px solid #000000">
<xsl:apply-templates select="address"/>
</td>
</tr>
</xsl:copy>
</xsl:template>
<!-- <xsl:for-each select="name">
<xsl:sort select="lname"/>
<xsl:value-of select="fname"/>
</xsl:for-each> -->
<xsl:template match="address">
<xsl:for-each select="line">
<xsl:apply-templates/>
<br/>
</xsl:for-each>
<xsl:value-of select="url"/>
<br/>
<xsl:value-of select="email"/>
<br/>
</xsl:template>
<xsl:template match="name">
<xsl:apply-templates/>
<xsl:text> </xsl:text>
</xsl:template>
</xsl:stylesheet>
====================================
file1.xml
====================================
<client>
<subclient>
<title>A</title>
</subclient>
<subclient>
<name>
<fname>Jack</fname>
<lname>John</lname>
</name>
<phones>
<phone/>
<ext/>
<fax/>
</phones>
<address>
<line>address 1</line>
<line>address 2</line>
<url>www.what.com</url>
<email/></address>
</subclient>
</client>
===================================
masterfile.xml
===================================
<masterfile>
<doc filename="file1.xml"/>
<doc filename="file2.xml"/>
</masterfile>
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------------------------------
Date: Thu, 30 Jun 2005 09:37:02 +0100
To: xsl-list(_at_)lists(_dot_)mulberrytech(_dot_)com
From: David Carlisle <davidc(_at_)nag(_dot_)co(_dot_)uk>
Subject: Re: [xsl] sorting issue
Message-Id:
<200506300837(_dot_)JAA23207(_at_)penguin(_dot_)nag(_dot_)co(_dot_)uk>
<xsl:preserve-space elements="text"/>
that line isn't doing anything (preserve space is the default)
<xsl:apply-templates
select="document(doc/@filename)/client">
This is just selecting one node from each document
which is why each document is sorted as a whole, you want t(I think) to
select all the subclients from all documents so that is
<xsl:apply-templates
select="document(doc/@filename)/client/subclient">
and then sort them by title
<xsl:sort select="name/lname" />
(some of your subclients have a title not an name currently they'll all
sort together as "" you may want to have title|name/lname or something
I don't konw...
David
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Date: Thu, 30 Jun 2005 08:37:54 +0200
To: <xsl-list(_at_)lists(_dot_)mulberrytech(_dot_)com>
From: "Huditsch Roman" <Roman(_dot_)Huditsch(_at_)lexisnexis(_dot_)at>
Subject: Problems grouping adjacent siblings
Message-ID:
<AA7BAD89B2B0C746A26452F44939F1E301603DE1(_at_)LNGVIEEX1(_dot_)orac(_dot_)local>
Good morning list,
I would like to group adjacent siblings which share the same local-name
into a container element.
Given this structure
<ax>
<e></e>
<r></r>
<exp>Example</exp>
<exp>bla bla</exp>
<exp>jada jada</exp>
<b></b>
<e></e>
<exp>Example 2</exp>
<exp>bla bla</exp>
<v></v>
</ax>
<bx>
<c></c>
<exp>Example 3</exp>
<exp>bla bla</exp>
<b></b>
</bx>
The following output should be produced:
<ax>
<e></e>
<r></r>
<remark>
<exp>Example</exp>
<exp>bla bla</exp>
<exp>jada jada</exp>
</remark>
<b></b>
<e></e>
<remark>
<exp>Example 2</exp>
<exp>bla bla</exp>
</remark>
<v></v>
</ax>
<bx>
<c></c>
<r></r>
<remark> =20
<exp>Example 3</exp>
<exp>bla bla</exp>
</remark>
<b></b>
</bx>
I thought that this would be the classical use-case for an
<xsl:for-each-group> with @group-adjacent, but I am getting the
strangest results...
I don't even know if it is better to do the grouping in a template
matching
exp[not(preceding-sibling::*[1][local-name()=3D'exp']) or in a template
matching exp's parent
I also tried to find a @group-starting-with solution, since the first
<exp> always starts with the string 'Example'. But with no success
either....
Can you help me with this problem?
Thank you very much for your help!
Wbr,
Roman
------------------------------
Date: Thu, 30 Jun 2005 01:06:17 -0700 (PDT)
To: xsl-list(_at_)lists(_dot_)mulberrytech(_dot_)com
From: Mukul Gandhi <mukul_gandhi(_at_)yahoo(_dot_)com>
Subject: Re: [xsl] Problems grouping adjacent siblings
Message-ID:
<20050630080617(_dot_)29379(_dot_)qmail(_at_)web32906(_dot_)mail(_dot_)mud(_dot_)yahoo(_dot_)com>
Hi Roman,
I thought the following stylesheet would have
worked.. But its not working. Its identity template
based ..
<?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" />
<xsl:template match="node() | @*">
<xsl:copy>
<xsl:apply-templates select="node() | @*" />
</xsl:copy>
</xsl:template>
<xsl:template match="exp">
<xsl:if test="not(name(preceding-sibling::*[1]) =
'exp')">
<remark>
<xsl:copy-of select=". | following-sibling::exp"
/>
</remark>
</xsl:if>
</xsl:template>
</xsl:stylesheet>
Regards,
Mukul
--- Huditsch Roman <Roman(_dot_)Huditsch(_at_)lexisnexis(_dot_)at>
wrote:
Good morning list,
I would like to group adjacent siblings which share
the same local-name
into a container element.
Given this structure
<ax>
<e></e>
<r></r>
<exp>Example</exp>
<exp>bla bla</exp>
<exp>jada jada</exp>
<b></b>
<e></e>
<exp>Example 2</exp>
<exp>bla bla</exp>
<v></v>
</ax>
<bx>
<c></c>
<exp>Example 3</exp>
<exp>bla bla</exp>
<b></b>
</bx>
The following output should be produced:
<ax>
<e></e>
<r></r>
<remark>
<exp>Example</exp>
<exp>bla bla</exp>
<exp>jada jada</exp>
</remark>
<b></b>
<e></e>
<remark>
<exp>Example 2</exp>
<exp>bla bla</exp>
</remark>
<v></v>
</ax>
<bx>
<c></c>
<r></r>
<remark>
<exp>Example 3</exp>
<exp>bla bla</exp>
</remark>
<b></b>
</bx>
I thought that this would be the classical use-case
for an
<xsl:for-each-group> with @group-adjacent, but I am
getting the
strangest results...
I don't even know if it is better to do the grouping
in a template
matching
exp[not(preceding-sibling::*[1][local-name()='exp'])
or in a template
matching exp's parent
I also tried to find a @group-starting-with
solution, since the first
<exp> always starts with the string 'Example'. But
with no success
either....
Can you help me with this problem?
Thank you very much for your help!
Wbr,
Roman
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Date: Thu, 30 Jun 2005 09:39:22 +0100
To: xsl-list(_at_)lists(_dot_)mulberrytech(_dot_)com
From: David Carlisle <davidc(_at_)nag(_dot_)co(_dot_)uk>
Subject: Re: [xsl] Problems grouping adjacent siblings
Message-Id:
<200506300839(_dot_)JAA23221(_at_)penguin(_dot_)nag(_dot_)co(_dot_)uk>
You want to use group-adjacent, grouping on name() no need to use the
sibling axis explictly here.
David
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------------------------------
Date: Thu, 30 Jun 2005 10:57:17 +0200
To: <xsl-list(_at_)lists(_dot_)mulberrytech(_dot_)com>
From: "Huditsch Roman" <Roman(_dot_)Huditsch(_at_)lexisnexis(_dot_)at>
Subject: AW: [xsl] Problems grouping adjacent siblings
Message-ID:
<AA7BAD89B2B0C746A26452F44939F1E3A54884(_at_)LNGVIEEX1(_dot_)orac(_dot_)local>
Hi
With the help of a colleague, I managed to get a working solution.
<xsl:template match=3D"*[exp]">
<xsl:copy>
<xsl:for-each-group select=3D"*" group-adjacent=3D"if (self::exp) =
then 'remark' else 'other'">
<xsl:choose>
<xsl:when test=3D"current-grouping-key() =3D 'other'">
<xsl:copy-of select=3D"current-group()"/>
</xsl:when>
<xsl:otherwise>
<xsl:element name=3D"{current-grouping-key()}">
<xsl:copy-of select=3D"current-group()"/>
</xsl:element>
</xsl:otherwise>
</xsl:choose>
</xsl:for-each-group>
</xsl:copy>
</xsl:template>
Thanks for the help!
wbr,
Roman
-----Urspr=FCngliche Nachricht-----
Von: David Carlisle [mailto:davidc(_at_)nag(_dot_)co(_dot_)uk]=20
Gesendet: Donnerstag, 30. Juni 2005 10:39
An: xsl-list(_at_)lists(_dot_)mulberrytech(_dot_)com
Betreff: Re: [xsl] Problems grouping adjacent siblings
=20
=20
You want to use group-adjacent, grouping on name() no need to=20
use the sibling axis explictly here.
=20
David
=20
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This e-mail has been scanned for all viruses by Star. The=20
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proactive anti-virus service working around the clock, around=20
the globe, visit:
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------------------------------
Date: Thu, 30 Jun 2005 18:01:27 +0100
To: <xsl-list(_at_)lists(_dot_)mulberrytech(_dot_)com>
From: "Michael Kay" <mike(_at_)saxonica(_dot_)com>
Subject: RE: [xsl] Problems grouping adjacent siblings
I would like to group adjacent siblings which share the same
local-name
into a container element.
I thought that this would be the classical use-case for an
<xsl:for-each-group> with @group-adjacent, but I am getting the
strangest results...
I can't tell what you are doing wrong without seeing your code.
With ax as the context node, you can do
<xsl:for-each-group select="*" group-adjacent="local-name()">
<xsl:choose>
<xsl:when test="current-grouping-key()='ex'">
<remark><xsl:copy-of select="current-group()"/></remark>
</
<xsl:otherwise>
<xsl:copy-of select="current-group()"/>
</
</
</
Michael Kay
http://www.saxonica.com/
------------------------------
Date: Thu, 30 Jun 2005 09:31:55 +0100 (BST)
To: xsl-list(_at_)lists(_dot_)mulberrytech(_dot_)com
From: Rahil Qamar <qamar_rahil(_at_)yahoo(_dot_)co(_dot_)uk>
Subject: Re: [xsl] problems with copying duplicate nodes
Message-ID:
<20050630083155(_dot_)69005(_dot_)qmail(_at_)web25502(_dot_)mail(_dot_)ukl(_dot_)yahoo(_dot_)com>
Thanks Mukul and David for the clarifications you
provided.
Much appreciate your help.
Rahil
--- Mukul Gandhi <mukul_gandhi(_at_)yahoo(_dot_)com> wrote:
Hi Rahil,
--- Rahil Qamar <qamar_rahil(_at_)yahoo(_dot_)co(_dot_)uk> wrote:
<xsl:variable name="rtf">
<temp>
<xsl:copy-of select="TD" />
</temp>
</xsl:variable>
So the variable 'rtf' contained the entire TD
content.
The variable rtf (Result Tree Fragment) contains a
XML
fragment, which is part of the Result Tree ..
<xsl:for-each select="$rtf/temp/TD[generate-id() =
generate-id(key('by-td', .)[1])]">
<xsl:copy-of select="." />
</xsl:for-each>
Then for each TD in the $rtf you checked whether
there
was a duplicate TD present bz comparing it with
the
first occurrence of TD.
Yes, thats true ..
Well I must thank you for providing me with a
solution. I had truly been struggling ! :)
I thank you for the nice problem you posted ..
Regards,
Mukul
Thanks
Rahil
--- Mukul Gandhi <mukul_gandhi(_at_)yahoo(_dot_)com> wrote:
Hi Rahil,
Please read my thoughts below your sentences
..
You
posted an interesting problem for me to find an
answer
.. Thanks..
--- Rahil Qamar <qamar_rahil(_at_)yahoo(_dot_)co(_dot_)uk> wrote:
(I am glad I was able to help ..)
The solution you posted worked even with my
Saxon
8.1.1 processor using XSLT 2.0.
Im getting the output just as I wanted but Im
intrigued about the functionality of the
<temp>
node
and how it eliminated the duplicates at a
later
stage.
I used temp as a convenience element, using
which
I
was able to distinguish element hierarchy
properly
..
We could write the stylesheet using without it
also
..
But I am not sure at this moment, how we should
do
that? I think David answered the question to
this
some
time back .. Please ask him :)
Just so I know how to resolve a similar query
in
future could you please explain what you did
in
the
<template match="TABLE/TR/TR"> block ?
In this template, I matched a pattern with a
portion
of the result tree (which was body of the
template)
..
I think the explanation could be as simple as
that
;)
Regards,
Mukul
Thanks again
Rahil
--- Mukul Gandhi <mukul_gandhi(_at_)yahoo(_dot_)com>
wrote:
Hi David,
Please read my response below your
comments
..
--- David Carlisle <davidc(_at_)nag(_dot_)co(_dot_)uk> wrote:
The stylesheet uses node-set function,
and
Muenchian
Grouping (over a RTF) ..
It _needs_ to use a node-set function, but
as
posted
you did not use it
(you need common:node-set($rtf))
Sorry if I am wrong. I tested the
stylesheet,
and
it
ran successfully with Saxon.
The stylesheet is tested with Saxon 8.4.
It's best not to test xslt 1 stylesheets
with
an
xslt2 processor.
As (as in this case) many things that
generate
an
error in xslt1 will
not be reported.
You are always right :)
Regards,
Mukul
David
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------------------------------
Date: Thu, 30 Jun 2005 11:05:06 +0200
To: xsl-list(_at_)lists(_dot_)mulberrytech(_dot_)com
From: James Fuller <jim(_dot_)fuller(_at_)ruminate(_dot_)co(_dot_)uk>
CC: ElektonikaMail(_at_)dollar(_dot_)fourthought(_dot_)com,
exslt(_at_)lists(_dot_)fourthought(_dot_)com
Subject: Re: [xsl] Re: [exslt] Re: [xsl] Importing compiled stylesheets
Message-ID: <42C3B5C2(_dot_)5010508(_at_)ruminate(_dot_)co(_dot_)uk>
Colin Paul Adams wrote:
Mike, what you say is true, but is not relevant to the case.
xsl:use-when IS a compile-time construction - it runs at compile-time,
not at run-time.
and the condiotion is not a runt-ime condition - element-available
evaluates, AT COMPILE TIME, whether or not the instruction is
available.
being very lazy at the moment,
is this behavior what the spec explicitly states or is it implementator
intepretation...also is it consistent between XSLT 1.0 and XSLT 2.0 ?
cheers, Jim Fuller
------------------------------
Date: 30 Jun 2005 10:48:17 +0100
To: xsl-list(_at_)lists(_dot_)mulberrytech(_dot_)com
From: Colin Paul Adams <colin(_at_)colina(_dot_)demon(_dot_)co(_dot_)uk>
Cc: ElektonikaMail(_at_)dollar(_dot_)fourthought(_dot_)com,
exslt(_at_)lists(_dot_)fourthought(_dot_)com
Subject: Re: [xsl] Re: [exslt] Re: [xsl] Importing compiled stylesheets
Message-ID: <m3ll4s41lq(_dot_)fsf(_at_)colina(_dot_)demon(_dot_)co(_dot_)uk>
"James" == James Fuller <jim(_dot_)fuller(_at_)ruminate(_dot_)co(_dot_)uk>
writes:
James> Colin Paul Adams wrote:
>>>>>>> Mike, what you say is true, but is not relevant to the
>>>>>>> case. xsl:use-when IS a compile-time construction - it
>>>>>>> runs at compile-time, not at run-time. and the condiotion
>>>>>>> is not a runt-ime condition - element-available evaluates,
>>>>>>> AT COMPILE TIME, whether or not the instruction is
>>>>>>> available.
>>>>>>>
>>>>>>>
James> being very lazy at the moment,
James> is this behavior what the spec explicitly states or is it
James> implementator intepretation...also is it consistent between
James> XSLT 1.0 and XSLT 2.0 ?
It is explicit, and it is not optional.
And it is not consistent, as XSLT 1.0 does not have a conditional
compilation facility.